NDA Maths · Trigonometric Equations

General Solutions & Counting Solutions

Because sine, cosine, and tangent repeat, a trig equation has an infinite family of solutions captured by one general formula — and counting how many fall in a given interval is the most-tested skill in the chapter.

All Trigonometric Equations notes NDA Maths strategy

Why this matters

The chapter's hardest pocket (13 PYQs, 6 HARD). Two skills dominate: writing the general solution after reducing the equation to a standard sin = sin / cos = cos / tan = tan form, and counting the solutions inside an interval. Get the reduction right and the counting follows.

Concept 1 of 4

The General-Solution Formulas

Intuition

Solving sin θ = sin α doesn't give one answer — it gives every angle co-terminal with α or its supplement. The three general-solution formulas package that whole infinite family into a single expression with an integer parameter n.

Definition

Write each equation in the form (function of θ) = (same function of a known angle α), then:

$\sin\theta = \sin\alpha \Rightarrow \theta = n\pi + (-1)^n \alpha$
$\cos\theta = \cos\alpha \Rightarrow \theta = 2n\pi \pm \alpha$
$\tan\theta = \tan\alpha \Rightarrow \theta = n\pi + \alpha$

for $n \in \mathbb{Z}$ . The principal solution is the one in the first cycle; the general solution adds the periodic family. (Special: $\sin\theta=0 \Rightarrow \theta=n\pi$ ; $\cos\theta=0 \Rightarrow \theta=(2n+1)\tfrac{\pi}{2}$ .)

General solutions

\sin\theta=\sin\alpha:\ \theta=n\pi+(-1)^n\alpha; \quad \cos\theta=\cos\alpha:\ \theta=2n\pi\pm\alpha; \quad \tan\theta=\tan\alpha:\ \theta=n\pi+\alpha; \quad \sin\theta=0:\ \theta=n\pi; \quad \cos\theta=0:\ \theta=(2n+1)\tfrac{\pi}{2}; \quad \sin^2\theta=\sin^2\alpha:\ \theta=n\pi\pm\alpha

Worked example

Find the general solution of

\cos\theta = \tfrac{1}{2}

Practice this concept6 quick reps

From the bank · past-year question

Example 1Trigonometric EquationsMODERATE

What is the general solution of

\cos^{100}x-\sin^{100}x=1

? (where

n

is an integer)

[Q27 · Sep · 2024]

sin and cos use DIFFERENT general forms

\sin\theta=\sin\alpha

uses

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

;

\cos\theta=\cos\alpha

uses

2n\pi\pm\alpha

. Swapping them is the most common error and changes which solutions you count.

The $(-1)^n$ belongs to SINE only

The alternating factor

(-1)^n

appears ONLY in the sine general solution

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

. Cosine uses

2n\pi\pm\alpha

and tangent uses

n\pi+\alpha

— neither carries a

(-1)^n

. Writing

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

is wrong.

The $\pm$ is the whole point of the cosine form

\cos\theta=\cos\alpha

gives BOTH branches

2n\pi+\alpha

and

2n\pi-\alpha

, so the formula is

2n\pi\pm\alpha

. Dropping the

\pm

and writing only

2n\pi+\alpha

loses half the solutions.

tan and sin share $n\pi$ but differ in the add-on

\tan\theta=\tan\alpha

gives

\theta=n\pi+\alpha

— note it is

n\pi

(period

\pi

), NOT

2n\pi

. Tangent repeats every

\pi

, so its general solution steps by

\pi

, unlike cosine's

2n\pi

Concept 2 of 4

Reducing an Equation to Standard Form

Intuition

Most equations don't arrive as sin = sin. The work is to massage them — co-function shifts, double-angle identities, or factoring a quadratic in one ratio — into a standard form. Watch for extraneous roots when you square.

Definition

Common reductions:

Co-function: $\sin 2\theta = \cos 3\theta = \sin(\tfrac{\pi}{2} - 3\theta)$ , then equate angles.
Quadratic in one ratio: equations like $2\cos^2 x + \cos x - 1 = 0$ factor; solve each linear factor.
Half-angle / Weierstrass: $\csc x + \cot x = \sqrt3$ becomes $1 + \cos x = \sqrt3\sin x$ ; square carefully.
Always verify roots in the ORIGINAL equation — squaring or dividing can introduce or drop solutions (e.g. where a denominator vanishes).

Co-function reduction

\cos\theta = \sin\!\left(\tfrac{\pi}{2} - \theta\right), \quad \sin\theta = \cos\!\left(\tfrac{\pi}{2} - \theta\right)

Worked example

Solve

\sin 3\theta = \cos 2\theta

for

0 < \theta < \tfrac{\pi}{2}

From the bank · past-year question

Example 2Trigonometric EquationsHARD

What is/are the solution(s) of the trigonometric equation

\csc x + \cot x = \sqrt{3}

, where

0 < x < 2\pi

[Q42 · Sep · 2018]

Squaring can add false roots

1 + \cos x = \sqrt3\sin x

squared gives a quadratic whose roots include

\cos x = -1

— which makes the original

\csc x + \cot x

undefined. Substitute every root back before counting.

Dividing by $\cos x$ (or $\sin x$ ) can LOSE roots

Going from

\sin x = \cos x

\tan x = 1

by dividing by

\cos x

silently assumes

\cos x \ne 0

. That is fine here, but in general dividing by a trig factor throws away every solution where that factor is zero — factor instead of divide.

$\sin x = 2$ has NO solution

After reducing, always check the range:

\sin x

and

\cos x

live in

[-1,1]

. An equation that forces

\sin x = 2

(or

\cos x = -3

) has no real solution at all — don't write a general solution for it.

Drill 5 more on reducing an equation to standard form

Concept 3 of 4

Counting Solutions in an Interval

Intuition

'How many solutions?' is asking how many members of the infinite family land inside the interval. Reduce to a single function = constant, then count the periods that fit — graphically, how many times the curve crosses the level line.

Definition

To count solutions of (function) $= k$ on an interval:

Reduce to one trig function equal to a constant (e.g. $\cot 2x\cot 3x = 1 \Rightarrow \cos 5x = 0$ ).
Find the general solution, then list the values of $n$ that keep $\theta$ inside the interval.
Discard values where the original equation is undefined (a cotangent/cosecant blowing up, a denominator zero).
Graphically, the count is the number of intersections of $y = (\text{function})$ with the horizontal line $y = k$ over the interval.

Count = solutions of the reduced equation in range

\cot 2x\,\cot 3x = 1 \ \Rightarrow\ \cos 5x = 0 \ \Rightarrow\ 5x = (2n+1)\tfrac{\pi}{2}

Worked example

How many solutions does

\sin 2x = \tfrac{1}{2}

have on

0 \le x < 2\pi

Practice this concept4 quick reps

From the bank · past-year question

Example 3Trigonometric EquationsMODERATE

What is the number of solutions of the equation

\cot 2x\cdot\cot 3x=1

for

0<x<\pi

[Q26 · Sep · 2024]

Scale the interval when the angle is multiplied

For

\sin 2x = k

0 \le x < 2\pi

, substitute

u = 2x

u

runs over

[0, 4\pi)

— TWICE the length. Counting solutions in the original

x

-interval instead of the stretched

u

-interval halves the count.

Discard solutions where the equation is undefined

When a reduced equation like

\cos 5x = 0

came from

\cot 2x\,\cot 3x = 1

, drop any value of

x

where

\cot 2x

\cot 3x

blows up — those are not genuine solutions of the original, even though they satisfy the reduced form.

Drill 4 more on counting solutions in an interval

Concept 4 of 4

Range & Existence Conditions

Intuition

Before solving, check whether a solution can exist at all: sine and cosine are trapped in [−1, 1], so an equation that forces a value outside that has no solution. The same bound limits how many parameter values are allowed.

Definition

$\sin x, \cos x \in [-1, 1]$ , so $a\sin x = b$ needs $|b| \le |a|$ ; count integer parameters by intersecting with this range.
For $\cos^{100}x - \sin^{100}x = 1$ : since both terms are in $[0,1]$ , equality forces $\cos^2 x = 1$ , i.e. $x = n\pi$ .
A boundary condition ( $0 \le x \le \tfrac{\pi}{2}$ ) plus an undefined-point check often cuts the candidate solutions down to one or two.

Existence bound

a\sin x = b \ \text{ solvable} \iff |b| \le |a|

Worked example

For how many integers

k

does

3\cos x = k

have a solution?

Practice this concept2 quick reps

From the bank · past-year question

Example 4Trigonometric EquationsMODERATE

The number of integer values of

k

, for which the equation

2\sin x = 2k+1

has a solution, is

[Q6 · Apr · 2021]

Count integers INCLUSIVELY across the range

For

3\cos x = k

the bound is

k \in [-3,3]

, which contains

7

integers

-3,\dots,3

— not

6

. Endpoints

\pm 3

are achievable (at

\cos x = \pm 1

), so include them; forgetting one or both endpoints is the classic off-by-one.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

The General-Solution Formulas
General solutions
$\sin\theta=\sin\alpha:\ \theta=n\pi+(-1)^n\alpha; \quad \cos\theta=\cos\alpha:\ \theta=2n\pi\pm\alpha; \quad \tan\theta=\tan\alpha:\ \theta=n\pi+\alpha; \quad \sin\theta=0:\ \theta=n\pi; \quad \cos\theta=0:\ \theta=(2n+1)\tfrac{\pi}{2}; \quad \sin^2\theta=\sin^2\alpha:\ \theta=n\pi\pm\alpha$
Reducing an Equation to Standard Form
Co-function reduction
$\cos\theta = \sin\!\left(\tfrac{\pi}{2} - \theta\right), \quad \sin\theta = \cos\!\left(\tfrac{\pi}{2} - \theta\right)$
Counting Solutions in an Interval
Count = solutions of the reduced equation in range
$\cot 2x\,\cot 3x = 1 \ \Rightarrow\ \cos 5x = 0 \ \Rightarrow\ 5x = (2n+1)\tfrac{\pi}{2}$
Range & Existence Conditions
Existence bound
$a\sin x = b \ \text{ solvable} \iff |b| \le |a|$

Watch out for (10)

sin and cos use DIFFERENT general forms
→ The General-Solution Formulas
The $(-1)^n$ belongs to SINE only
→ The General-Solution Formulas
The $\pm$ is the whole point of the cosine form
→ The General-Solution Formulas
tan and sin share $n\pi$ but differ in the add-on
→ The General-Solution Formulas
Squaring can add false roots
→ Reducing an Equation to Standard Form
Dividing by $\cos x$ (or $\sin x$ ) can LOSE roots
→ Reducing an Equation to Standard Form
$\sin x = 2$ has NO solution
→ Reducing an Equation to Standard Form
Scale the interval when the angle is multiplied
→ Counting Solutions in an Interval
Discard solutions where the equation is undefined
→ Counting Solutions in an Interval
Count integers INCLUSIVELY across the range
→ Range & Existence Conditions

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