MHT-CET Maths · Trigonometric Functions

Trigonometric Equations — General Solutions, Counting Roots and a cos x + b sin x

A trigonometric equation is solved by reducing it to one ratio equal to one value, writing the general solution from three fixed patterns, and then keeping only the roots the stated interval and the original equation allow.

All Trigonometric Functions notes MHT-CET Maths strategy

Why this matters

47 PYQs, 36% HARD. The paper asks the same five things in rotation: a general or principal solution, a quadratic in one ratio with a root to reject, a cos x + b sin x = c, an equation solved by factorising a sum of sines, and a range argument that shows an equation has no solution or counts its roots. The marks are lost in the last step — keeping a root where the original equation is undefined, or counting over the wrong interval — so every concept here ends with a check.

Concept 1 of 5: General and Principal Solutions of sin θ = k, cos θ = k, tan θ = k

A trigonometric ratio repeats, so an equation like

\sin\theta = \frac12

has infinitely many solutions. Find ONE angle that works (the principal value

\alpha

), then add the repeat: tangent repeats every

\pi

, cosine is symmetric about 0, and sine is symmetric about

\frac{\pi}{2}

, which is where the

(-1)^n

comes from.

Definition

$\sin\theta = \sin\alpha \Rightarrow \theta = n\pi + (-1)^n\alpha$ .
$\cos\theta = \cos\alpha \Rightarrow \theta = 2n\pi \pm \alpha$ .
$\tan\theta = \tan\alpha \Rightarrow \theta = n\pi + \alpha$ , with $n \in \mathbb{Z}$ throughout.
Principal solutions are the values in $[0, 2\pi)$ . Find the reference angle, then place it by the quadrant signs: for $\cos x = -\frac{\sqrt3}{2}$ , the reference angle is $\frac{\pi}{6}$ and cosine is negative in quadrants II and III, giving $\frac{5\pi}{6}$ and $\frac{7\pi}{6}$ .
Two conditions at once (for example $\sin\theta < 0$ and $\tan\theta > 0$ ) fix the quadrant; take the one angle that satisfies both.
Convert first: $\cot\theta = \tan\left(\frac{\pi}{2} - \theta\right)$ , $\cos x = \sin\left(\frac{\pi}{2} - x\right)$ , and $\dfrac{\tan x - 1}{\tan x + 1} = \tan\left(x - \frac{\pi}{4}\right)$ . An equation between two different ratios becomes one pattern after this.

The three general-solution patterns

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

$\alpha$ any one solution — usually the principal value
nany integer

Worked example

Solve

\tan 2\theta = \cot\theta

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 11th May Shift 2 · Q108Hard

Example 1 · Trigonometric Functions · Trigonometric Equations and General Solutions

If the general solution of the equation

\frac{\tan 3x - 1}{\tan 3x + 1} = \sqrt{3}

\frac{n\pi}{p} + \frac{7\pi}{q},\ n,p,q \in \mathbb{Z}

, then

pq

Using the sine pattern for cosine

The

(-1)^n

belongs to SINE only.

\cos\theta = \cos\alpha

gives

2n\pi \pm \alpha

;

\tan

gives

n\pi + \alpha

. The options always include the wrong pattern for the right

\alpha

Drill 9 more on general and principal solutions of sin θ = k, cos θ = k, tan θ = k

Concept 2 of 5: Equations That Reduce to a Quadratic in One Ratio — and the Roots to Reject

Use

\sin^2 + \cos^2 = 1

to rewrite everything in one ratio; the equation becomes an ordinary quadratic. Then two checks remove wrong answers: a root outside

[-1, 1]

for sine or cosine is impossible, and a root where the ORIGINAL equation divides by zero is not a solution even though the algebra produced it.

Definition

Replace $\cos^2 x$ by $1 - \sin^2 x$ (or the reverse) so only one ratio remains; factorise.
Reject $\sin x$ or $\cos x$ values outside $[-1, 1]$ : $3\sin^2 x - 7\sin x + 2 = 0$ gives $\sin x = \frac13$ or $2$ , and only $\frac13$ survives.
Reject roots where the equation is undefined. $\tan x + \sec x = 2\cos x$ becomes $2\sin^2 x + \sin x - 1 = 0$ , so $\sin x = \frac12$ or $-1$ ; but $\sin x = -1$ means $\cos x = 0$ , where $\tan x$ and $\sec x$ do not exist. Only $\frac{\pi}{6}$ and $\frac{5\pi}{6}$ remain.
Counting in an interval: $\sin x = k$ with $0 < k < 1$ has two roots per full turn, both in the half-turns where sine is positive. In $(0, 5\pi)$ those half-turns are $(0, \pi)$ , $(2\pi, 3\pi)$ , $(4\pi, 5\pi)$ : six roots.

Reduce, then check

\cos^2 x = 1-\sin^2 x \qquad -1 \le \sin x,\ \cos x \le 1 \qquad \tan x,\ \sec x \text{ need } \cos x \ne 0

Worked example

Solve

2\cos^2 x + 3\sin x = 3

[0, 2\pi]

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 9th May Shift 1 · Q141Hard

Example 2 · Trigonometric Functions · Trigonometric Equations and General Solutions

(0,2\pi)

, the number of solutions of

\tan\theta+\sec\theta=2\cos\theta

are

Counting the root the equation cannot hold

When the equation contains

\tan

\sec

\cot

\csc

, clearing denominators can create a root where one of them is undefined. The option that counts it (3 instead of 2 for

\tan x + \sec x = 2\cos x

) is always there.

Drill 10 more on equations that reduce to a quadratic in one ratio — and the roots to reject

Concept 3 of 5: a cos x + b sin x = c — the R Form, When a Solution Exists, and Roots as a Pair

a\cos x + b\sin x

is a single wave in disguise: it equals

R\cos(x - \phi)

with

R = \sqrt{a^2 + b^2}

. So it can never exceed

R

in size, which answers every 'for which k does a solution exist' question, and once written as one cosine it solves by the ordinary pattern.

Definition

$a\cos x + b\sin x = R\cos(x - \phi)$ where $R = \sqrt{a^2 + b^2}$ , $\cos\phi = \frac{a}{R}$ , $\sin\phi = \frac{b}{R}$ .
Existence: $a\cos x + b\sin x = c$ has a solution exactly when $|c| \le \sqrt{a^2 + b^2}$ . The range of $a\cos x + b\sin x$ is $[-R, R]$ .
Divide by $R$ to solve: $\sin x + \cos x = 1 \Rightarrow \cos\left(x - \frac{\pi}{4}\right) = \frac{1}{\sqrt2}$ , so $x = 2n\pi$ or $2n\pi + \frac{\pi}{2}$ .
Roots as a pair: with $t = \tan x$ , $\cos 2x = \frac{1 - t^2}{1 + t^2}$ and $\sin 2x = \frac{2t}{1 + t^2}$ turn $a\cos 2x + b\sin 2x = c$ into a quadratic in $t$ . Its roots are $\tan\alpha, \tan\beta$ , so Vieta gives $\tan\alpha + \tan\beta$ and $\tan\alpha\tan\beta$ , and $\tan(\alpha + \beta)$ follows from the compound-angle formula.

The R form and the existence condition

a\cos x+b\sin x=\sqrt{a^2+b^2}\,\cos(x-\phi) \qquad \text{solvable} \iff |c|\le\sqrt{a^2+b^2}

$\phi$ the angle with $\cos\phi = a/R$ , $\sin\phi = b/R$

Worked example

For how many integers

k

does

3\cos x + 4\sin x = k - 1

have a solution?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 1 · Q131Moderate

Example 3 · Trigonometric Functions · Trigonometric Equations and General Solutions

The number of integral values of

k

for which the equation

7\cos x+5\sin x=2k+1

has a solution, is

Counting the integers at the boundary

\sqrt{74} \approx 8.6

is not an integer, so

-8.6 \le 2k + 1 \le 8.6

gives

-4.8 \le k \le 3.8

: the integers are

-4

3

, eight of them. Rounding the bound before solving for

k

is the usual way to lose one.

Drill 6 more on a cos x + b sin x = c — the r form, when a solution exists, and roots as a pair

Concept 4 of 5: Solving by Factorisation — Sum-to-Product and Multiple Angles

An equation with three or more angles is almost never solved term by term. Pair the outer terms with sum-to-product so a common factor appears, then set each factor to zero — every factor is now a one-ratio equation of the first concept.

Definition

$\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$ , $\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$ .
Pair the terms whose average is the middle angle: in $\sin\theta + \sin 4\theta + \sin 7\theta$ , pair $\sin 7\theta + \sin\theta = 2\sin 4\theta\cos 3\theta$ , so the whole is $\sin 4\theta(2\cos 3\theta + 1)$ .
Multiple angles reduce to one ratio: $\sin 2\theta = 2\sin\theta\cos\theta$ , $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ . Divide out a factor only after noting when it is zero.
Count each factor separately in the interval, then check no root is shared by two factors.
A factor that can never vanish (such as $2\cos x - 3$ ) contributes nothing, and the question usually says so in its condition.

Sum-to-product

\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2} \qquad \cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}

Worked example

Solve

\cos x + \cos 3x = \cos 2x

\left(0, \frac{\pi}{2}\right)

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 10th May Shift 1 · Q123Hard

Example 4 · Trigonometric Functions · Trigonometric Equations and General Solutions

Number of solutions of

\sin\theta+\sin4\theta+\sin7\theta=0,\;\theta\in(0,\pi)

Dividing by a factor that can be zero

Cancelling

\sin 3x

from both sides of

2\sin 3x\cos 2x = \sin 3x

loses every root of

\sin 3x = 0

. Move everything to one side and factor instead.

Drill 7 more on solving by factorisation — sum-to-product and multiple angles

Concept 5 of 5: Range Arguments — No Solution, Exponential Forms and Trigonometric Inequalities

Some equations are settled before any solving: if one side can only take values in a range the other side never reaches, there is no solution. The same bounds turn exponential equations like

16^{\sin^2 x} + 16^{\cos^2 x} = 10

into a quadratic with the answer read off from

0 \le \sin^2 x \le 1

Definition

Bounds to use: $-1 \le \sin x, \cos x \le 1$ ; $\frac1e \le e^{\sin x} \le e$ ; $\frac12 \le \sin^4 x + \cos^4 x \le 1$ (since it equals $1 - \frac12\sin^2 2x$ ).
No solution: $e^{\sin x} - e^{-\sin x} = 4$ needs $e^{\sin x} \approx 4.24$ , but $e^{\sin x} \le e \approx 2.72$ .
Exponential forms: in $a^{\sin^2 x} + a^{\cos^2 x} = k$ , put $y = a^{\sin^2 x}$ ; then $a^{\cos^2 x} = \frac{a}{y}$ and $y + \frac{a}{y} = k$ is a quadratic. Each root fixes $\sin^2 x$ , and each value of $\sin^2 x$ strictly between 0 and 1 gives four roots in $[0, 2\pi]$ .
Inequalities: solve the trigonometric part for its interval ( $2\sin^2 x + 3\sin x - 2 > 0 \Rightarrow \sin x > \frac12$ ), solve the algebraic part, then intersect on a number line using $\frac{\pi}{6} \approx 0.52$ , $\frac{5\pi}{6} \approx 2.62$ .

Bounds that decide an equation

\sin^4 x+\cos^4 x = 1-\tfrac12\sin^2 2x \in \left[\tfrac12,\,1\right] \qquad a^{\sin^2 x}\cdot a^{\cos^2 x}=a

Worked example

How many solutions does

9^{\sin^2 x} + 9^{\cos^2 x} = 10

have in

[0, \pi]

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 1 · Q130Hard

Example 5 · Trigonometric Functions · Trigonometric Equations and General Solutions

The number of solutions in

[0,2\pi]

of the equation

16^{\sin^2 x}+16^{\cos^2 x}=10

Counting two roots per value of sin²x instead of four

\sin^2 x = \frac14

means

\sin x = \frac12

-\frac12

, and each has two roots in

[0, 2\pi]

. Over

[0, \pi]

only the positive value counts, and it has two.

A second root of an equation the key counts once

(1 - \tan^2\theta)\sec^2\theta + 2^{\tan^2\theta} = 0

is keyed 2 values, from

\tan^2\theta = 3

. The equation

2^t = t^2 - 1

also crosses near

t \approx 3.41

. The paper's answer is 2; know that it counts only the exact root.

Drill 10 more on range arguments — no solution, exponential forms and trigonometric inequalities

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 (5)

General and Principal Solutions of sin θ = k, cos θ = k, tan θ = k
The three general-solution patterns
$\sin\theta=\sin\alpha \Rightarrow \theta=n\pi+(-1)^n\alpha \qquad \cos\theta=\cos\alpha \Rightarrow \theta=2n\pi\pm\alpha \qquad \tan\theta=\tan\alpha \Rightarrow \theta=n\pi+\alpha$
Equations That Reduce to a Quadratic in One Ratio — and the Roots to Reject
Reduce, then check
$\cos^2 x = 1-\sin^2 x \qquad -1 \le \sin x,\ \cos x \le 1 \qquad \tan x,\ \sec x \text{ need } \cos x \ne 0$
a cos x + b sin x = c — the R Form, When a Solution Exists, and Roots as a Pair
The R form and the existence condition
$a\cos x+b\sin x=\sqrt{a^2+b^2}\,\cos(x-\phi) \qquad \text{solvable} \iff |c|\le\sqrt{a^2+b^2}$
Solving by Factorisation — Sum-to-Product and Multiple Angles
Sum-to-product
$\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2} \qquad \cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$
Range Arguments — No Solution, Exponential Forms and Trigonometric Inequalities
Bounds that decide an equation
$\sin^4 x+\cos^4 x = 1-\tfrac12\sin^2 2x \in \left[\tfrac12,\,1\right] \qquad a^{\sin^2 x}\cdot a^{\cos^2 x}=a$

Watch out for (6)

Using the sine pattern for cosine
→ General and Principal Solutions of sin θ = k, cos θ = k, tan θ = k
Counting the root the equation cannot hold
→ Equations That Reduce to a Quadratic in One Ratio — and the Roots to Reject
Counting the integers at the boundary
→ a cos x + b sin x = c — the R Form, When a Solution Exists, and Roots as a Pair
Dividing by a factor that can be zero
→ Solving by Factorisation — Sum-to-Product and Multiple Angles
Counting two roots per value of sin²x instead of four
→ Range Arguments — No Solution, Exponential Forms and Trigonometric Inequalities
A second root of an equation the key counts once
→ Range Arguments — No Solution, Exponential Forms and Trigonometric Inequalities

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