Trigonometric equations and general values | NEB Class 11 Mathematics Notes

Chapter Overview

This chapter is the bridge between basic trigonometry and advanced calculus. We learn to solve Trigonometric Equations — equations involving trigonometric functions of unknown angles (e.g., $\sin \theta = 1/2$). Because trig functions are periodic (values repeat after certain intervals), a single equation can have infinite solutions. This chapter teaches us how to represent all those infinite solutions using a single expression called the General Solution.

What this chapter covers

Principal vs General SolutionsPeriodicity of Trigonometric FunctionsGeneral Solutions for sin, cos, and tanEquations with Zero ValuesSquared Equations (sin², cos², tan²)Factoring Method for Quadratic TypeMultiple Angle EquationsEquations of the form a sin x + b cos x = cCD Formula Grouping MethodUsing tan(x/2) Substitution

Learning Objectives

1

Distinguish between principal solutions (in [0, 2π)) and general solutions (involving integer n)

2

Solve basic trigonometric equations for sin, cos, and tan using standard general formulas

3

Solve complex equations using factoring, quadratic substitution, and transformation formulas

4

Master the method for equations of the type a sin x + b cos x = c by dividing by √(a²+b²)

5

Apply C-D (sum-to-product) formulas to group and factor equations with multiple angles

6

Recognise question types instantly and choose the correct solving strategy

The One Idea Behind This Entire Chapter

Key Insight

Unlike algebraic equations (like x² − 4 = 0) which have a finite number of solutions, trigonometric equations have infinite solutions because sin, cos, and tan repeat their values periodically.

The general solution is a single formula with an integer n that captures every possible answer at once. Your job is: identify the equation type → match it to the correct formula → write the general solution with n ∈ ℤ.

Example: sin θ = 0 has solutions at 0, π, 2π, 3π, ... and also −π, −2π, ... The general solution θ = nπ captures ALL of them in one line.

Key Definitions

Trigonometric Equation

An equation involving trigonometric functions of an unknown angle. Examples: $\sin \theta = 1/2$, $\cos 2x = \sqrt{3}/2$, $\tan^2 x = 3$.

Solution

The value of the unknown angle that satisfies the given trigonometric equation.

Principal Solution

The solutions that lie in the interval [0, 2π) — the smallest non-negative solutions. There are a finite number of principal solutions.

$$\theta \in [0, 2\pi)$$

General Solution

A generalized expression involving an integer 'n' that represents all possible solutions due to the periodicity of the trig function.

$$\text{General solution includes } n \in \mathbb{Z}$$

Periodicity

The property that a function's values repeat after a fixed interval. sin and cos have period 2π; tan has period π. This is why trig equations have infinite solutions.

Principal Solution vs General Solution

The most important distinction in this chapter. Understand when each is asked and how to find them.

Principal Solution

$$\theta \in [0, 2\pi)$$

Solutions in the interval [0, 2π) only

Finite number of solutions (usually 1 or 2)

No 'n' involved — just specific angle values

Find using the unit circle or special angles table

Keywords: 'find the principal value', 'find the least positive value'

VS

General Solution

$$\theta = f(n), \; n \in \mathbb{Z}$$

Represents ALL possible solutions (infinite)

Contains an integer n where n ∈ ℤ

Uses the periodicity of trig functions

Each formula type (sin, cos, tan, squared) has its own pattern

Keywords: 'general solution', 'general value', 'solve completely'

Memory trick: Principal = particular values in [0, 2π). General = gives every solution with n.

Formula Sheet — All General Solutions with Derivations

sin θ = 0

\theta = n\pi

sin is zero at all integer multiples of π.

1sin θ = 0 at θ = 0, π, 2π, 3π, ... and also at negative multiples: −π, −2π, ... All these values share one pattern: they are integer multiples of π.

2Therefore, the compact general solution is: $\theta = n\pi, \quad n \in \mathbb{Z}$. Every single solution to sin θ = 0 is captured by this one formula.

cos θ = 0

\theta = (2n+1)\frac{\pi}{2}

cos is zero at all odd multiples of π/2: π/2, 3π/2, 5π/2, ...

1cos θ = 0 at θ = π/2, 3π/2, 5π/2, ... and also −π/2, −3π/2, ... These are all odd multiples of π/2.

2An odd number can be written as (2n+1) where n is any integer. Therefore: $\theta = (2n+1)\frac{\pi}{2}, \quad n \in \mathbb{Z}$.

tan θ = 0

\theta = n\pi

tan = sin/cos, so tan is zero whenever sin is zero (and cos ≠ 0).

1Since $\tan \theta = \frac{\sin \theta}{\cos \theta}$, tan θ = 0 whenever sin θ = 0 (provided cos θ ≠ 0).

2Therefore, tan θ = 0 has the same general solution as sin θ = 0: $\theta = n\pi, \quad n \in \mathbb{Z}$.

sin θ = sin α

\theta = n\pi + (-1)^n \alpha

The (−1)ⁿ factor makes the sign switch — plus when n is even, minus when n is odd.

1Step 1: Write the equation as a difference: $\sin \theta - \sin \alpha = 0$

2Step 2: Apply the C-D formula: $\sin C - \sin D = 2\cos\frac{C+D}{2} \sin\frac{C-D}{2}$. This gives: $2\cos\left(\frac{\theta+\alpha}{2}\right) \sin\left(\frac{\theta-\alpha}{2}\right) = 0$

3Step 3: Set each factor to zero: either $\cos\left(\frac{\theta+\alpha}{2}\right) = 0$ or $\sin\left(\frac{\theta-\alpha}{2}\right) = 0$.

4Step 4: Solve each case using basic zero-formulas and simplify to get: $$\theta = n\pi + (-1)^n \alpha, \quad n \in \mathbb{Z}$$

cos θ = cos α

\theta = 2n\pi \pm \alpha

Cosine is 'Consistent' — always uses 2nπ and the ± sign.

1Step 1: Write as difference: $\cos \theta - \cos \alpha = 0$

2Step 2: Use C-D formula: $\cos C - \cos D = -2\sin\frac{C+D}{2}\sin\frac{C-D}{2}$. This gives: $-2\sin\left(\frac{\theta+\alpha}{2}\right) \sin\left(\frac{\theta-\alpha}{2}\right) = 0$

3Step 3: Set each factor to zero and use $\sin x = 0 \Rightarrow x = n\pi$:
$\frac{\theta+\alpha}{2} = n\pi \Rightarrow \theta = 2n\pi - \alpha$ AND $\frac{\theta-\alpha}{2} = n\pi \Rightarrow \theta = 2n\pi + \alpha$.

4Step 4: Combine both cases: $$\theta = 2n\pi \pm \alpha, \quad n \in \mathbb{Z}$$

tan θ = tan α

\theta = n\pi + \alpha

Simplest of the three — just add nπ to the reference angle.

1Step 1: Write as: $\tan \theta - \tan \alpha = 0$. Express in sin/cos: $\frac{\sin \theta}{\cos \theta} - \frac{\sin \alpha}{\cos \alpha} = 0$.

2Step 2: Combine fractions: $\frac{\sin\theta\cos\alpha - \cos\theta\sin\alpha}{\cos\theta\cos\alpha} = 0$. The numerator is $\sin(\theta - \alpha)$.

3Step 3: Set $\sin(\theta - \alpha) = 0 \Rightarrow \theta - \alpha = n\pi$. Therefore: $$\theta = n\pi + \alpha, \quad n \in \mathbb{Z}$$

sin² θ = sin² α (also cos², tan²)

\theta = n\pi \pm \alpha

ALL squared equations (sin², cos², tan²) give θ = nπ ± α. No exceptions!

1sin² case: Use $\sin^2 \theta = \frac{1-\cos 2\theta}{2}$: both sides reduce to $\cos 2\theta = \cos 2\alpha$.

2Apply cos formula: $2\theta = 2n\pi \pm 2\alpha$. Divide by 2: $$\theta = n\pi \pm \alpha$$

3cos² and tan² follow the same logic: they all reduce to either $\cos 2\theta = \cos 2\alpha$ or $\sin^2 \theta = \sin^2 \alpha$, both giving the same $\theta = n\pi \pm \alpha$ result.

a sin x + b cos x = c

R = \sqrt{a^2+b^2}, \quad \text{check } |c| \le R \text{ first}

The most important 4-mark question type. Always divide by √(a²+b²) first. Equation has no solution if |c| > R.

1Step 1: Calculate $R = \sqrt{a^2 + b^2}$. Check: if $|c| > R$, write 'no solution' and stop.

2Step 2: Divide the entire equation by R: $\frac{a}{R}\sin x + \frac{b}{R}\cos x = \frac{c}{R}$.

3Step 3: Find angle α such that $\cos \alpha = \frac{a}{R}$ and $\sin \alpha = \frac{b}{R}$ (or matching pattern). Rewrite LHS as $\sin(x + \alpha)$ or $\cos(x \pm \alpha)$ using compound angle formula.

4Step 4: Solve the resulting simple trig equation using the standard general formulas.

Memory Tricks — Never Forget the Formulas

Trick: Three simple memory hooks cover every basic general solution formula on exam day.

Steps to Remember:

Sine is 'Switchy': The $(-1)^n$ in the sine formula makes the sign switch between plus and minus depending on whether n is even (+) or odd (−). Formula: $\theta = n\pi + (-1)^n \alpha$.
Cos is 'Consistent': It always uses $2n\pi$ and the $\pm$ sign. Formula: $\theta = 2n\pi \pm \alpha$. No switching — just always both plus and minus.
Squares are Simple: If the equation has squared trig functions ($\sin^2, \cos^2, \tan^2$), the answer is ALWAYS $\theta = n\pi \pm \alpha$. No exceptions. All three squared forms share the same general solution.

Mnemonic: Sine = Switchy [(−1)ⁿ] | Cos = Consistent [2nπ ± α] | Squares = Simple [nπ ± α]

The Factorisation Trick — Never Divide by sin x or cos x

Trick: When both sides have a common trig factor, NEVER cancel it by dividing — you will lose solutions. Always factorise.

Steps to Remember:

Wrong approach: $2\sin^2 x + \sin x = 0$ → dividing by sin x gives $2\sin x + 1 = 0$ → only gets part of the answer.
Correct approach: Factorise → $\sin x(2\sin x + 1) = 0$ → Case 1: $\sin x = 0$ → Case 2: $\sin x = -1/2$. Both cases give the complete solution.
Rule: If you see $\sin x \cdot (\text{something}) = 0$, factor it. Never divide. Every factor set to zero gives a family of solutions.

Mnemonic: Never divide. Always factorise. Each factor = one family of general solutions.

Common Mistakes

These mistakes account for the most lost marks in NEB exams for this chapter.

Forgetting to check the range: Not verifying if a solution like $\sin x = 2$ is possible. Remember: $|\sin x| \le 1$ and $|\cos x| \le 1$. If the RHS exceeds these bounds, write 'no solution' immediately.
Missing n ∈ ℤ: Always mention that n is an integer in your general solution. Without it, the answer is mathematically incomplete and loses marks.
Dividing by a variable trig function: Never divide both sides by $\sin x$ or $\cos x$ — you lose solutions. Bring everything to one side and factorise instead.
Forgetting ± in cos formula: $\cos \theta = \cos \alpha$ gives two families of solutions ($+\alpha$ and $-\alpha$). Don't just write one.
Confusing sin and cos formulas: sin uses $(-1)^n$ and $n\pi$. cos uses $2n\pi \pm \alpha$. Keep them separate.
Not checking a sin x + b cos x = c for validity: The equation has a solution only if $c^2 \le a^2 + b^2$. If $c > \sqrt{a^2+b^2}$, write 'no solution' immediately.

NEB Exam Traps — These Appear Every Year

NEB Exam Warning

Board examiners deliberately test these edge cases. Students who don't know them lose 1–4 marks per question.

$\sin x = k$ where $|k| > 1$ — students try to solve it instead of instantly writing 'no solution'. $|\sin x|$ can never exceed 1.
$\sec x = k$ where $|k| < 1$ — since $\sec x = 1/\cos x$ and $|\cos x| \le 1$, we need $|\sec x| \ge 1$. No solution if $|k| < 1$.
Dividing by $\sin x$ or $\cos x$ and losing roots — always factorise instead of dividing.
Forgetting the $(-1)^n$ in the sine general solution formula — the most commonly forgotten piece.
Questions of type $\tan x + \tan 2x = \tan 3x$ — use the $\tan(A+B)$ identity to simplify, don't solve each term separately.
For $a\sin x + b\cos x = c$: missing the range check $|c| \le \sqrt{a^2+b^2}$ before proceeding.

Question Recognition Guide

Keyword / Phrase in Question	Use This
sin θ = k (simple value)	Find α where sin α = k, then: θ = nπ + (−1)ⁿα
cos θ = k (simple value)	Find α where cos α = k, then: θ = 2nπ ± α
tan θ = k (simple value)	Find α where tan α = k, then: θ = nπ + α
sin² θ = k, cos² θ = k, tan² θ = k	ALL squared equations → θ = nπ ± α
Quadratic type: a sin²x + b sin x + c = 0	Substitute y = sin x, factorise like normal quadratic, solve each factor
Multiple angles: sin 3x = sin x, cos 4x = cos 2x	Apply standard formula directly with the multiple angles
a sin x + b cos x = c	Divide by √(a²+b²), use compound angle, convert to single sin or cos
sin A + sin B + sin C = 0 type	Group terms, use C-D (sum-to-product) formulas, then factorise
Equation has \|RHS\| > 1 for sin/cos	INSTANTLY write 'No solution' — check range first!
Finding principal solution only	Find values in [0, 2π) only — no n needed
Finding general solution	Must include n ∈ ℤ in the final answer

Standard Angle Reference Table

Memorise these values — every trig equation question uses these standard angles as reference angles α.

Angle (degrees)	Angle (radians)	sin	cos	tan
0°	0	0	1	0
30°	π/6	1/2	√3/2	1/√3
45°	π/4	1/√2	1/√2	1
60°	π/3	√3/2	1/2	√3
90°	π/2	1	0	undefined
120°	2π/3	√3/2	−1/2	−√3
135°	3π/4	1/√2	−1/√2	−1
150°	5π/6	1/2	−√3/2	−1/√3
180°	π	0	−1	0
270°	3π/2	−1	0	undefined
360°	2π	0	1	0

Solved Examples (Easy → Hard)

1

Find the general solution of $\cos \theta = -\frac{\sqrt{3}}{2}$.

Easy

Solution

Step 1: cos is negative in Quadrants II and III.

Step 2: We know $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$, so the reference angle is $\frac{\pi}{6}$.

Step 3: $-\frac{\sqrt{3}}{2} = \cos\left(\pi - \frac{\pi}{6}\right) = \cos \frac{5\pi}{6}$. So $\alpha = \frac{5\pi}{6}$.

Step 4: Apply $\cos \theta = \cos \alpha$ formula:
$$\theta = 2n\pi \pm \frac{5\pi}{6}, \quad n \in \mathbb{Z}$$

∴ θ = 2nπ ± 5π/6, n ∈ ℤ

2

Find the principal solutions of $\sin x = \frac{\sqrt{3}}{2}$.

Easy

Solution

sin is positive in Quadrants I and II. We need values in $[0, 2\pi)$.

Q-I: $x = \frac{\pi}{3}$ (since $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$).

Q-II: $x = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.

Both $\frac{\pi}{3}$ and $\frac{2\pi}{3}$ are in $[0, 2\pi)$.

∴ Principal solutions: x = π/3 and x = 2π/3

3

Solve $2\sin^2 x + \sin x = 0$.

Medium

Solution

Step 1 — Factorise: Don't divide by sin x (you'd lose solutions)!
$$\sin x(2\sin x + 1) = 0$$

Case 1: $\sin x = 0 \Rightarrow x = n\pi$

Case 2: $2\sin x + 1 = 0 \Rightarrow \sin x = -\frac{1}{2} = \sin\left(-\frac{\pi}{6}\right)$

Using $\sin x = \sin \alpha$ formula:
$$x = n\pi + (-1)^n\left(-\frac{\pi}{6}\right), \quad n \in \mathbb{Z}$$

∴ x = nπ OR x = nπ + (−1)ⁿ(−π/6), n ∈ ℤ

4

Solve $4\cos^2 \theta - 3 = 0$.

Medium

Solution

Step 1: Rearrange: $\cos^2 \theta = \frac{3}{4} = \left(\frac{\sqrt{3}}{2}\right)^2 = \cos^2 \frac{\pi}{6}$

Step 2: All squared trig equations use $\theta = n\pi \pm \alpha$:
$$\theta = n\pi \pm \frac{\pi}{6}, \quad n \in \mathbb{Z}$$

∴ θ = nπ ± π/6, n ∈ ℤ

5

Solve $\sqrt{3}\sin x - \cos x = \sqrt{2}$.

Hard

Solution

Step 1 — Check validity: $R = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{4} = 2$. Since $|\sqrt{2}| < 2$, solution exists.

Step 2 — Divide by R = 2:
$\frac{\sqrt{3}}{2}\sin x - \frac{1}{2}\cos x = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$

Step 3 — Recognise compound angle:
$\cos \frac{\pi}{6}\sin x - \sin \frac{\pi}{6}\cos x = \frac{1}{\sqrt{2}}$

This is $\sin\left(x - \frac{\pi}{6}\right) = \frac{1}{\sqrt{2}} = \sin \frac{\pi}{4}$

Step 4 — Apply sin formula:
$x - \frac{\pi}{6} = n\pi + (-1)^n \frac{\pi}{4}$

$$x = n\pi + (-1)^n \frac{\pi}{4} + \frac{\pi}{6}, \quad n \in \mathbb{Z}$$

∴ x = nπ + (−1)ⁿπ/4 + π/6, n ∈ ℤ

MCQ Practice

Mixed Level — MCQ

1

The general solution of $\sin x = 1$ is:

Correct!

Incorrect. Correct: $2n\pi + \frac{\pi}{2}$

Solution

$\sin x = 1$ occurs at $x = \frac{\pi}{2}$ in $[0, 2\pi)$.

Since $\sin x$ has period $2\pi$:
$$x = 2n\pi + \frac{\pi}{2}, \quad n \in \mathbb{Z}$$

∴ $2n\\pi + \\frac{\\pi}{2}$

2

If $\tan^2 \theta = 1$, the general value of $\theta$ is:

Correct!

Incorrect. Correct: $n\pi \pm \frac{\pi}{4}$

Solution

$\tan^2 \theta = 1 = \tan^2 \frac{\pi}{4}$

For ALL squared trig equations, use $\theta = n\pi \pm \alpha$:
$$\theta = n\pi \pm \frac{\pi}{4}, \quad n \in \mathbb{Z}$$

∴ $n\\pi \\pm \\frac{\\pi}{4}$

3

The number of principal solutions of $\sin x = \frac{1}{2}$ is:

Correct!

Incorrect. Correct: 2

Solution

In $[0, 2\pi)$:

Q-I: $x = \frac{\pi}{6}$ (since $\sin \frac{\pi}{6} = \frac{1}{2}$)
Q-II: $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$

Both in $[0, 2\pi)$. Number of principal solutions = 2

∴ 2

4

Solve $\cos 3\theta = 0$:

Correct!

Incorrect. Correct: $(2n+1)\frac{\pi}{6}$

Solution

Using $\cos x = 0 \Rightarrow x = (2n+1)\frac{\pi}{2}$:

$3\theta = (2n+1)\frac{\pi}{2}$

Divide by 3:
$$\theta = (2n+1)\frac{\pi}{6}, \quad n \in \mathbb{Z}$$

∴ $(2n+1)\\frac{\\pi}{6}$

5

Which of the following has no solution?

Correct!

Incorrect. Correct: $\sec x = 0.5$

Solution

$\sec x = \frac{1}{\cos x}$. Since $|\cos x| \le 1$, we have $|\sec x| \ge 1$.

$|\sec x| = 0.5 < 1$ is impossible.

No solution exists.

∴ $\\sec x = 0.5$

6

The general solution of $\sin x = \cos x$ is:

Correct!

Incorrect. Correct: $n\pi + \frac{\pi}{4}$

Solution

Divide both sides by $\cos x$:
$\tan x = 1 = \tan \frac{\pi}{4}$

Using $\tan x = \tan \alpha \Rightarrow x = n\pi + \alpha$:
$$x = n\pi + \frac{\pi}{4}, \quad n \in \mathbb{Z}$$

∴ $n\\pi + \\frac{\\pi}{4}$

7

General solution of $\sin^2 x = \frac{3}{4}$:

Correct!

Incorrect. Correct: $n\pi \pm \frac{\pi}{3}$

Solution

$\sin^2 x = \frac{3}{4} = \left(\frac{\sqrt{3}}{2}\right)^2 = \sin^2 \frac{\pi}{3}$

For ALL squared trig equations:
$$x = n\pi \pm \frac{\pi}{3}, \quad n \in \mathbb{Z}$$

∴ $n\\pi \\pm \\frac{\\pi}{3}$

8

If $\sin \theta = 0$ and $\cos \theta = -1$, then $\theta$ is:

Correct!

Incorrect. Correct: $(2n+1)\pi$

Solution

$\sin \theta = 0 \Rightarrow \theta = n\pi$.

At $\theta = n\pi$:
n even → $\cos(n\pi) = 1$ ✗
n odd → $\cos(n\pi) = -1$ ✓

For n odd, write $n = 2m+1$:
$$\theta = (2n+1)\pi, \quad n \in \mathbb{Z}$$

∴ $(2n+1)\\pi$

9

General solution of $\tan 2x = \tan x$:

Correct!

Incorrect. Correct: $n\pi$

Solution

Using $\tan A = \tan B \Rightarrow A = n\pi + B$:

$2x = n\pi + x$
$2x - x = n\pi$
$$x = n\pi, \quad n \in \mathbb{Z}$$

∴ $n\\pi$

10

The equation $\sqrt{3}\sin \theta + \cos \theta = 4$ has:

Correct!

Incorrect. Correct: No solution

Solution

For $a\sin\theta + b\cos\theta = c$, the range is $[-\sqrt{a^2+b^2},\; \sqrt{a^2+b^2}]$.

$R = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{4} = 2$

Since $c = 4 > R = 2$, the value exceeds the maximum possible.

No solution exists.

∴ No solution

Short Questions (2 Marks Each)

1

Find the principal solutions of $\tan x = -\sqrt{3}$.

[2 marks]

Solution

tan is negative in Q-II and Q-IV. Reference angle: $\tan \frac{\pi}{3} = \sqrt{3}$.

Q-II: $x = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$

Q-IV: $x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$

Both in $[0, 2\pi)$.

∴ x = 2π/3, 5π/3

2

Solve $4\cos^2 \theta = 3$.

[2 marks]

Solution

$\cos^2 \theta = \frac{3}{4} = \left(\frac{\sqrt{3}}{2}\right)^2 = \cos^2 \frac{\pi}{6}$

Squared equation → use $\theta = n\pi \pm \alpha$:
$$\theta = n\pi \pm \frac{\pi}{6}, \quad n \in \mathbb{Z}$$

∴ θ = nπ ± π/6, n ∈ ℤ

3

Find the general value of x if $\sin 2x = \sin x$.

[2 marks]

Solution

Using $\sin A = \sin B \Rightarrow A = n\pi + (-1)^n B$:
$2x = n\pi + (-1)^n x$

n even: $2x = 2m\pi + x \Rightarrow x = 2m\pi$

n odd: $2x = (2m+1)\pi - x \Rightarrow 3x = (2m+1)\pi \Rightarrow x = (2m+1)\frac{\pi}{3}$

∴ x = 2nπ OR x = (2n+1)π/3, n ∈ ℤ

4

Solve $\tan 3x + \tan x = 0$.

[2 marks]

Solution

$\tan 3x = -\tan x = \tan(-x)$

Using $\tan A = \tan B \Rightarrow A = n\pi + B$:
$3x = n\pi + (-x)$
$3x + x = n\pi$
$4x = n\pi$
$$x = \frac{n\pi}{4}, \quad n \in \mathbb{Z}$$

∴ x = nπ/4, n ∈ ℤ

5

Find the general solution of $\sec^2 \theta = 2$.

[2 marks]

Solution

$\sec^2 \theta = 2 \Rightarrow \cos^2 \theta = \frac{1}{2} = \left(\frac{1}{\sqrt{2}}\right)^2 = \cos^2 \frac{\pi}{4}$

Squared equation:
$$\theta = n\pi \pm \frac{\pi}{4}, \quad n \in \mathbb{Z}$$

∴ θ = nπ ± π/4, n ∈ ℤ

6

Solve $2\sin^2 x + 3\sin x + 1 = 0$.

[2 marks]

Solution

Let y = sin x:
$2y^2 + 3y + 1 = 0$
$(2y+1)(y+1) = 0$

$\sin x = -\frac{1}{2}$ or $\sin x = -1$

$\sin x = -\frac{1}{2} = \sin\left(-\frac{\pi}{6}\right) \Rightarrow x = n\pi + (-1)^n\left(-\frac{\pi}{6}\right)$

$\sin x = -1 = \sin\left(-\frac{\pi}{2}\right) \Rightarrow x = n\pi + (-1)^n\left(-\frac{\pi}{2}\right)$

∴ x = nπ + (−1)ⁿ(−π/6) OR x = nπ + (−1)ⁿ(−π/2)

7

Solve $\cot \theta = 0$.

[2 marks]

Solution

$\cot \theta = \frac{\cos \theta}{\sin \theta} = 0 \Rightarrow \cos \theta = 0$ (with $\sin \theta \neq 0$).

Using $\cos \theta = 0$ formula:
$$\theta = (2n+1)\frac{\pi}{2}, \quad n \in \mathbb{Z}$$

∴ θ = (2n+1)π/2, n ∈ ℤ

8

Solve $\tan x + \cot x = 2$.

[2 marks]

Solution

$\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = 2$

$\frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = 2$

$\frac{1}{\sin x \cos x} = 2 \Rightarrow \sin 2x = 1$

$2x = 2n\pi + \frac{\pi}{2} \Rightarrow x = n\pi + \frac{\pi}{4}$

∴ x = nπ + π/4, n ∈ ℤ

9

Find the general solution of $\sin 5x = 0$.

[2 marks]

Solution

Using $\sin \theta = 0 \Rightarrow \theta = n\pi$:

$5x = n\pi$

$$x = \frac{n\pi}{5}, \quad n \in \mathbb{Z}$$

∴ x = nπ/5, n ∈ ℤ

10

Solve $\cos x = \sin 2x$.

[2 marks]

Solution

Convert: $\sin 2x = \cos\left(\frac{\pi}{2} - 2x\right)$

$\cos x = \cos\left(\frac{\pi}{2} - 2x\right)$

Using $\cos A = \cos B \Rightarrow A = 2n\pi \pm B$:

Case (+): $x = 2n\pi + \frac{\pi}{2} - 2x \Rightarrow 3x = 2n\pi + \frac{\pi}{2} \Rightarrow x = \frac{2n\pi}{3} + \frac{\pi}{6}$

Case (−): $x = 2n\pi - \frac{\pi}{2} + 2x \Rightarrow -x = 2n\pi - \frac{\pi}{2} \Rightarrow x = \frac{\pi}{2} - 2n\pi$

∴ x = 2nπ/3 + π/6 OR x = π/2 − 2nπ

Long Questions (4 Marks Each)

1

Solve: $\sqrt{3}\sin x + \cos x = 1$.

[4 marks]

Solution

Step 1 — Check validity: $R = \sqrt{(\sqrt{3})^2 + 1^2} = 2$. Since $|1| < 2$, solution exists.

Step 2 — Divide by R = 2:
$\frac{\sqrt{3}}{2}\sin x + \frac{1}{2}\cos x = \frac{1}{2}$

Step 3 — Recognise compound angle:
$\cos \frac{\pi}{6}\sin x + \sin \frac{\pi}{6}\cos x = \frac{1}{2}$
$\sin\left(x + \frac{\pi}{6}\right) = \frac{1}{2} = \sin \frac{\pi}{6}$

Step 4 — Apply sin formula:
$x + \frac{\pi}{6} = n\pi + (-1)^n \frac{\pi}{6}$
$x = n\pi + (-1)^n \frac{\pi}{6} - \frac{\pi}{6}$

n even (n=2m): $x = 2m\pi + \frac{\pi}{6} - \frac{\pi}{6} = 2m\pi$

n odd (n=2m+1): $x = (2m+1)\pi - \frac{\pi}{6} - \frac{\pi}{6} = (2m+1)\pi - \frac{\pi}{3}$

∴ x = 2nπ OR x = (2n+1)π − π/3, n ∈ ℤ

2

Solve: $\sin \theta + \sin 2\theta + \sin 3\theta = 0$.

[4 marks]

Solution

Step 1 — Group terms:
$(\sin 3\theta + \sin \theta) + \sin 2\theta = 0$

Step 2 — Apply C-D formula $\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$:
$2\sin 2\theta \cos \theta + \sin 2\theta = 0$

Step 3 — Factorise:
$\sin 2\theta (2\cos \theta + 1) = 0$

Case 1: $\sin 2\theta = 0 \Rightarrow 2\theta = n\pi \Rightarrow \theta = \frac{n\pi}{2}$

Case 2: $2\cos \theta + 1 = 0 \Rightarrow \cos \theta = -\frac{1}{2} = \cos \frac{2\pi}{3}$
$$\theta = 2n\pi \pm \frac{2\pi}{3}$$

∴ θ = nπ/2 OR θ = 2nπ ± 2π/3, n ∈ ℤ

3

Solve: $4\sin^2 x - 8\cos x + 1 = 0$.

[4 marks]

Solution

Step 1 — Convert using $\sin^2 x = 1 - \cos^2 x$:
$4(1-\cos^2 x) - 8\cos x + 1 = 0$
$4\cos^2 x + 8\cos x - 5 = 0$

Step 2 — Factorise (let y = cos x):
$4y^2 + 8y - 5 = 0$
$4y^2 + 10y - 2y - 5 = 0$
$(2y-1)(2y+5) = 0$

Step 3: $y = \frac{1}{2}$ or $y = -\frac{5}{2}$
Reject $y = -\frac{5}{2}$ since $|\cos x| \le 1$.

Step 4: $\cos x = \frac{1}{2} = \cos \frac{\pi}{3}$:
$$x = 2n\pi \pm \frac{\pi}{3}, \quad n \in \mathbb{Z}$$

∴ x = 2nπ ± π/3, n ∈ ℤ

4

Solve: $\tan \theta + \tan 2\theta = \tan 3\theta$.

[4 marks]

Solution

Step 1 — Use the identity $\tan(A+B) = \frac{\tan A + \tan B}{1-\tan A \tan B}$:

$\tan 3\theta = \tan(\theta + 2\theta) = \frac{\tan \theta + \tan 2\theta}{1 - \tan \theta \tan 2\theta}$

Step 2 — Substitute into given equation:
$\tan \theta + \tan 2\theta = \frac{\tan \theta + \tan 2\theta}{1 - \tan \theta \tan 2\theta}$

Step 3 — Let $T = \tan \theta + \tan 2\theta$:
$T(1 - \tan \theta \tan 2\theta) = T$
$-T \cdot \tan \theta \tan 2\theta = 0$

Step 4: Either $T = 0$ or $\tan \theta = 0$ or $\tan 2\theta = 0$

Each case gives: $\theta = n\pi$ or $2\theta = n\pi$:
$$\theta = \frac{n\pi}{3}, \quad n \in \mathbb{Z}$$

∴ θ = nπ/3, n ∈ ℤ

5

Solve: $2\sin^2 x + \sin^2 2x = 2$.

[4 marks]

Solution

Step 1: Expand $\sin^2 2x = (2\sin x \cos x)^2 = 4\sin^2 x \cos^2 x$:
$2\sin^2 x + 4\sin^2 x \cos^2 x = 2$

Step 2 — Divide by 2:
$\sin^2 x + 2\sin^2 x \cos^2 x = 1$

Step 3 — Use $\cos^2 x = 1 - \sin^2 x$:
$\sin^2 x + 2\sin^2 x(1-\sin^2 x) = 1$
$2\sin^4 x - 3\sin^2 x + 1 = 0$

Step 4 — Let $u = \sin^2 x$:
$(2u-1)(u-1) = 0$

$\sin^2 x = \frac{1}{2} = \sin^2 \frac{\pi}{4} \Rightarrow x = n\pi \pm \frac{\pi}{4}$
$\sin^2 x = 1 = \sin^2 \frac{\pi}{2} \Rightarrow x = n\pi \pm \frac{\pi}{2}$

∴ x = nπ ± π/4 OR x = nπ ± π/2, n ∈ ℤ

6

Solve: $\cos x - \sin x = \frac{1}{\sqrt{2}}$.

[4 marks]

Solution

Step 1: $R = \sqrt{1^2 + (-1)^2} = \sqrt{2}$. Since $\frac{1}{\sqrt{2}} < \sqrt{2}$, solution exists.

Step 2 — Divide by $\sqrt{2}$:
$\frac{1}{\sqrt{2}}\cos x - \frac{1}{\sqrt{2}}\sin x = \frac{1}{2}$

Step 3 — Recognise compound angle:
$\cos \frac{\pi}{4}\cos x - \sin \frac{\pi}{4}\sin x = \frac{1}{2}$
$\cos\left(x + \frac{\pi}{4}\right) = \frac{1}{2} = \cos \frac{\pi}{3}$

Step 4 — Apply cos formula:
$x + \frac{\pi}{4} = 2n\pi \pm \frac{\pi}{3}$
$x = 2n\pi \pm \frac{\pi}{3} - \frac{\pi}{4}$

$x = 2n\pi + \frac{\pi}{12}$ or $x = 2n\pi - \frac{7\pi}{12}$

∴ x = 2nπ + π/12 OR x = 2nπ − 7π/12, n ∈ ℤ

7

Solve: $\sin \theta - \sin 3\theta + \sin 5\theta = 0$.

[4 marks]

Solution

Step 1 — Group: $(\sin 5\theta + \sin \theta) - \sin 3\theta = 0$

Step 2 — C-D formula: $\sin 5\theta + \sin \theta = 2\sin 3\theta \cos 2\theta$
$2\sin 3\theta \cos 2\theta - \sin 3\theta = 0$

Step 3 — Factorise:
$\sin 3\theta(2\cos 2\theta - 1) = 0$

Case 1: $\sin 3\theta = 0 \Rightarrow 3\theta = n\pi \Rightarrow \theta = \frac{n\pi}{3}$

Case 2: $\cos 2\theta = \frac{1}{2} = \cos \frac{\pi}{3}$
$2\theta = 2n\pi \pm \frac{\pi}{3} \Rightarrow \theta = n\pi \pm \frac{\pi}{6}$

∴ θ = nπ/3 OR θ = nπ ± π/6, n ∈ ℤ

8

Solve: $\tan^2 x + \sec x = 1$.

[4 marks]

Solution

Step 1 — Use $\tan^2 x = \sec^2 x - 1$:
$(\sec^2 x - 1) + \sec x = 1$
$\sec^2 x + \sec x - 2 = 0$

Step 2 — Factorise (let y = sec x):
$(y+2)(y-1) = 0$
$\sec x = -2$ or $\sec x = 1$

Case 1: $\sec x = 1 \Rightarrow \cos x = 1 \Rightarrow x = 2n\pi$

Case 2: $\sec x = -2 \Rightarrow \cos x = -\frac{1}{2} = \cos \frac{2\pi}{3}$
$$x = 2n\pi \pm \frac{2\pi}{3}$$

∴ x = 2nπ OR x = 2nπ ± 2π/3, n ∈ ℤ

9

Solve: $12\cos^3 x - 3\cos x = 0$.

[4 marks]

Solution

Step 1 — Factorise:
$3\cos x(4\cos^2 x - 1) = 0$

Case 1: $\cos x = 0$
$$x = (2n+1)\frac{\pi}{2}$$

Case 2: $4\cos^2 x - 1 = 0 \Rightarrow \cos^2 x = \frac{1}{4} = \cos^2 \frac{\pi}{3}$
$$x = n\pi \pm \frac{\pi}{3}$$

∴ x = (2n+1)π/2 OR x = nπ ± π/3, n ∈ ℤ

10

Solve: $\sin 2x \tan x + 1 = \sin 2x + \tan x$.

[4 marks]

Solution

Step 1 — Rearrange:
$\sin 2x \tan x - \sin 2x - \tan x + 1 = 0$

Step 2 — Factor by grouping:
$\sin 2x(\tan x - 1) - 1(\tan x - 1) = 0$
$(\sin 2x - 1)(\tan x - 1) = 0$

Case 1: $\sin 2x = 1 \Rightarrow 2x = 2n\pi + \frac{\pi}{2} \Rightarrow x = n\pi + \frac{\pi}{4}$

Case 2: $\tan x = 1 = \tan \frac{\pi}{4} \Rightarrow x = n\pi + \frac{\pi}{4}$

Both cases give the same solution set.

∴ x = nπ + π/4, n ∈ ℤ

Past NEB Board Questions

1

Solve: $\sin x + \sin 2x + \sin 3x = 0$.

NEB 2079 [4 marks]

Solution

Group: $(\sin 3x + \sin x) + \sin 2x = 0$

C-D formula: $2\sin 2x \cos x + \sin 2x = 0$

Factorise: $\sin 2x(2\cos x + 1) = 0$

Case 1: $\sin 2x = 0 \Rightarrow x = \frac{n\pi}{2}$

Case 2: $\cos x = -\frac{1}{2} = \cos \frac{2\pi}{3} \Rightarrow x = 2n\pi \pm \frac{2\pi}{3}$

∴ x = nπ/2 OR x = 2nπ ± 2π/3, n ∈ ℤ

2

Solve: $\sqrt{3}\sin x + \cos x = \sqrt{2}$.

NEB 2078 [4 marks]

Solution

R = $\sqrt{3+1} = 2$. Since $|\sqrt{2}| < 2$, solution exists.

Divide by 2: $\frac{\sqrt{3}}{2}\sin x + \frac{1}{2}\cos x = \frac{\sqrt{2}}{2}$

$\cos \frac{\pi}{6}\sin x + \sin \frac{\pi}{6}\cos x = \frac{1}{\sqrt{2}}$

$\sin\left(x + \frac{\pi}{6}\right) = \frac{1}{\sqrt{2}} = \sin \frac{\pi}{4}$

$x + \frac{\pi}{6} = n\pi + (-1)^n \frac{\pi}{4}$
$$x = n\pi + (-1)^n \frac{\pi}{4} - \frac{\pi}{6}$$

∴ x = nπ + (−1)ⁿπ/4 − π/6, n ∈ ℤ

3

Solve: $4\sin^2 x - 8\cos x + 1 = 0$.

NEB 2077 [4 marks]

Solution

Use $\sin^2 x = 1 - \cos^2 x$:
$4 - 4\cos^2 x - 8\cos x + 1 = 0$
$4\cos^2 x + 8\cos x - 5 = 0$
$(2\cos x - 1)(2\cos x + 5) = 0$

$\cos x = \frac{1}{2}$ (accept) or $\cos x = -\frac{5}{2}$ (reject, out of range)

$\cos x = \cos \frac{\pi}{3}$:
$$x = 2n\pi \pm \frac{\pi}{3}$$

∴ x = 2nπ ± π/3, n ∈ ℤ

4

Find the general solution of $\sin 2x + \sin 4x + \sin 6x = 0$.

NEB 2076 [4 marks]

Solution

Group: $(\sin 6x + \sin 2x) + \sin 4x = 0$

C-D formula: $2\sin 4x \cos 2x + \sin 4x = 0$

Factorise: $\sin 4x(2\cos 2x + 1) = 0$

Case 1: $\sin 4x = 0 \Rightarrow 4x = n\pi \Rightarrow x = \frac{n\pi}{4}$

Case 2: $\cos 2x = -\frac{1}{2} = \cos \frac{2\pi}{3}$
$2x = 2n\pi \pm \frac{2\pi}{3} \Rightarrow x = n\pi \pm \frac{\pi}{3}$

∴ x = nπ/4 OR x = nπ ± π/3, n ∈ ℤ

Chapter Test — Full Mixed Paper

Attempt this test in 45 minutes without looking at notes.

1 Section A — MCQs (1 Mark Each)

Mixed Level — MCQ

1

General solution of $\tan \theta = 0$ is:

Correct!

Incorrect. Correct: $n\pi$

Solution

$\tan \theta = 0 \Rightarrow \theta = n\pi, \quad n \in \mathbb{Z}$

∴ $n\\pi$

2

$\cos \theta = 1 \Rightarrow \theta = ?$

Correct!

Incorrect. Correct: $2n\pi$

Solution

$\cos \theta = 1 = \cos 0$, so $\theta = 2n\pi \pm 0 = 2n\pi$

∴ $2n\\pi$

3

Solution of $\sqrt{2}\sin x = 1$ is:

Correct!

Incorrect. Correct: $n\pi + (-1)^n \frac{\pi}{4}$

Solution

$\sin x = \frac{1}{\sqrt{2}} = \sin \frac{\pi}{4}$
$$x = n\pi + (-1)^n \frac{\pi}{4}, \quad n \in \mathbb{Z}$$

∴ $n\\pi + (-1)^n \\frac{\\pi}{4}$

2 Section B — Short Questions (2 Marks Each)

1

Solve $\cos 4x = \cos 2x$.

[2 marks]

Solution

Using $\cos A = \cos B \Rightarrow A = 2n\pi \pm B$:

Case (+): $4x = 2n\pi + 2x \Rightarrow x = n\pi$

Case (−): $4x = 2n\pi - 2x \Rightarrow 6x = 2n\pi \Rightarrow x = \frac{n\pi}{3}$

∴ x = nπ OR x = nπ/3

2

Solve $\tan^2 \theta = 3$.

[2 marks]

Solution

$\tan^2 \theta = 3 = (\sqrt{3})^2 = \tan^2 \frac{\pi}{3}$

For squared equations: $\theta = n\pi \pm \frac{\pi}{3}$

∴ θ = nπ ± π/3, n ∈ ℤ

3 Section C — Long Questions (4 Marks Each)

1

Solve $\sqrt{3}\sin x + \cos x = \sqrt{2}$.

[4 marks]

Solution

Divide by $\sqrt{3+1}=2$: $\frac{\sqrt{3}}{2}\sin x + \frac{1}{2}\cos x = \frac{\sqrt{2}}{2}$

$\sin(x + \frac{\pi}{6}) = \frac{1}{\sqrt{2}} = \sin \frac{\pi}{4}$

$x = n\pi + (-1)^n \frac{\pi}{4} - \frac{\pi}{6}$

∴ x = nπ + (−1)ⁿπ/4 − π/6, n ∈ ℤ

2

Solve $\sin 2x + \sin 4x + \sin 6x = 0$.

[4 marks]

Solution

Group: $(\sin 6x + \sin 2x) + \sin 4x = 0$

C-D: $2\sin 4x \cos 2x + \sin 4x = 0$

Factorise: $\sin 4x(2\cos 2x + 1) = 0$

$\sin 4x = 0 \Rightarrow x = \frac{n\pi}{4}$

$\cos 2x = -\frac{1}{2} = \cos \frac{2\pi}{3} \Rightarrow x = n\pi \pm \frac{\pi}{3}$

∴ x = nπ/4 OR x = nπ ± π/3

One-Page Revision Cheat Sheet

1

Principal Value: Smallest positive angle in [0, 2π). No 'n' needed.

2

General Value: Add periodicity factor with n ∈ ℤ.

3

sin θ = sin α → θ = nπ + (−1)ⁿα (Sine is Switchy)

4

cos θ = cos α → θ = 2nπ ± α (Cos is Consistent)

5

tan θ = tan α → θ = nπ + α (Tan is Simple — just add nπ)

6

ALL squared equations (sin², cos², tan²) → θ = nπ ± α — no exceptions

7

Zero formulas: sin θ = 0 → nπ; cos θ = 0 → (2n+1)π/2; tan θ = 0 → nπ

8

Range check FIRST: Is |RHS| ≤ 1 for sin/cos? If not → No solution

9

a sin x + b cos x = c: Check |c| ≤ √(a²+b²) first. Divide by R = √(a²+b²), use compound angle

10

Never divide by sin x or cos x — always factorise instead

11

Multiple angles (sin 3x, cos 4x): solve for the whole multiple, then divide

12

Grouping method: sin A + sin B = 2sin((A+B)/2)cos((A-B)/2)

13

Always write n ∈ ℤ after every general solution — mandatory for full marks

14

C-D formula for sums: group outer terms, apply C-D, factorise, solve each factor