Trigonometric Equations

Q: What is the general solution?

The general solution is a formula involving an integer n that gives every solution of the equation. E.g. sin θ = sin α has general solution θ = nπ + (-1)ⁿ α, n ∈ ℤ.

Q: What is the general solution of sin θ = sin α?

θ = nπ + (-1)ⁿ α, where α ∈ [-π/2, π/2] and n ∈ ℤ. The alternating (-1)ⁿ handles the two solutions per period.

Q: What is the general solution of cos θ = cos α?

θ = 2nπ ± α, where α ∈ [0, π] and n ∈ ℤ. The ± captures the fact that cosine is even, so both α and -α satisfy the equation.

Q: What is the general solution of tan θ = tan α?

θ = nπ + α, where α ∈ (-π/2, π/2) and n ∈ ℤ. Tangent has period π, so a single nπ shift captures every solution.

Q: How do I solve a sin x + b cos x = c?

Divide both sides by √(a² + b²) to get sin(x + φ) = c/√(a² + b²) (where tan φ = b/a). A solution exists if and only if |c| ≤ √(a² + b²).

Q: Why do I get extra solutions when I square a trig equation?

Squaring can turn a false statement into a true one (e.g., -1 = 1 is false, but (-1)^2 = 1^2 is true). Every squaring step may introduce extraneous solutions - so verify each candidate in the original equation.

Q: Can a trigonometric equation have no solution?

Yes. For example sin x = 2 has no real solution because sin x ∈ [-1, 1]. Similarly sin x + cos x = 3 has none because the maximum of the LHS is √2 < 3.

MathsTrigonometryFor JEE aspirants

A trigonometric equation is an equation involving trigonometric ratios of an unknown angle. Because trig functions are periodic, most such equations have infinitely many solutions - captured by a general solution written in terms of an integer $n$ . This page covers principal solutions, general solutions of the six standard forms $(sin θ = sin α, cos θ = cos α, tan θ = tan α$ , and their squared versions $)$ , and the seven standard solving techniques (factorisation, quadratic reduction, sum-to-product, product-to-sum, $a sin x + b cos x = c$ , half-angle substitution, and boundedness). Essential for JEE Main and JEE Advanced.

General Solutions - Quick Reference

$sin θ = sin α \Rightarrow θ = nπ + (- 1)^{n} α$ , $α \in [- π /2, π /2]$
$cos θ = cos α \Rightarrow θ = 2 nπ \pm α$ , $α \in [0, π]$
$tan θ = tan α \Rightarrow θ = nπ + α$ , $α \in (- π /2, π /2)$
$sin^{2} θ = sin^{2} α \Rightarrow θ = nπ \pm α$ ; same for $cos^{2}, tan^{2}$
$sin θ = 0 \Rightarrow θ = nπ$ ; $cos θ = 0 \Rightarrow θ = (2 n + 1) π /2$ ; $tan θ = 0 \Rightarrow θ = nπ$
$sin θ = 1 \Rightarrow θ = (4 n + 1) π /2$ ; $cos θ = 1 \Rightarrow θ = 2 nπ$ ; $cos θ = - 1 \Rightarrow θ = (2 n + 1) π$

Everywhere $n \in Z$ .

1. Trigonometric Equation and its Solution

An equation involving trigonometric ratios of an unknown angle is called a trigonometric equation. A solution is a value of the unknown that satisfies the equation.

For example, $sin θ = 1/ 2$ has solutions $θ = π /4, 3 π /4, 9 π /4, 11 π /4, \dots$ - infinitely many, because sine is periodic. The solutions split naturally into two categories:

Principal solutions: those lying in $[0, 2 π)$ .
General solution: a formula involving an integer $n$ that captures all solutions at once.

1.1 Principal Solutions - an Example

Solved Example 1

Find the principal solutions of

sin x = 1/2

Solution:

Looking for $x \in [0, 2 π)$ with $sin x = 1/2$ . From standard values, $sin (π /6) = 1/2$ . Sine is also positive in quadrant II with the same reference angle, so $sin (5 π /6) = 1/2$ . Both lie in $[0, 2 π)$ .

$x = π /6 or x = 5 π /6$

Figure: Principal solutions of

sin x = 1/2

are

π /6

and

5 π /6

2. General Solutions of Standard Equations

(i)

sin θ = sin α \Rightarrow θ = nπ + (- 1)^{n} α

, where

α \in [- π /2, π /2]

n \in Z

.

(ii)

cos θ = cos α \Rightarrow θ = 2 nπ \pm α

, where

α \in [0, π]

n \in Z

.

(iii)

tan θ = tan α \Rightarrow θ = nπ + α

, where

α \in (- π /2, π /2)

n \in Z

2.1 Squared Equations

sin^{2} θ = sin^{2} α

cos^{2} θ = cos^{2} α

tan^{2} θ = tan^{2} α

all give

θ = nπ \pm α, n \in Z

The angle

α

is called the principal angle of the equation.

2.2 Common Special Cases

Equation	General solution
$sin θ = 0$	$θ = nπ$
$sin θ = 1$	$θ = (4 n + 1) \frac{π}{2}$
$sin θ = - 1$	$θ = (4 n - 1) \frac{π}{2}$
$cos θ = 0$	$θ = (2 n + 1) \frac{π}{2}$
$cos θ = 1$	$θ = 2 nπ$
$cos θ = - 1$	$θ = (2 n + 1) π$
$tan θ = 0$	$θ = nπ$

Solved Example 2

Solve

sin θ = 3 /2

Solution:

$sin θ = 3 /2 = sin (π /3)$ . General: $θ = nπ + (- 1)^{n} \frac{π}{3}$ , $n \in Z$ .

Solved Example 3

Solve

sec 2 θ = - 2/ 3

Solution:

$sec 2 θ = - 2/ 3 \Rightarrow cos 2 θ = - 3 /2 = cos (5 π /6)$ .

General: $2 θ = 2 nπ \pm 5 π /6 \Rightarrow θ = nπ \pm 5 π /12$ , $n \in Z$ .

Solved Example 4

Solve

tan θ = 2

Solution:

Set $2 = tan α$ where $α = tan^{- 1} (2)$ . Then $θ = nπ + tan^{- 1} (2)$ , $n \in Z$ .

Solved Example 5

Solve

cos^{2} θ = 1/2

Solution:

$cos^{2} θ = (1/ 2)^{2} = cos^{2} (π /4)$ . So $θ = nπ \pm π /4$ , $n \in Z$ .

Solved Example 6

Solve

4 tan^{2} θ = 3 sec^{2} θ

Solution:

Domain: $cos θ \neq = 0$ . Rewrite $4 sin^{2} θ / cos^{2} θ = 3/ cos^{2} θ$ , i.e. $4 sin^{2} θ = 3$ (multiplying both sides by $cos^{2} θ$ , valid since it is nonzero).

$sin^{2} θ = 3/4 = (3 /2)^{2} = sin^{2} (π /3) \Rightarrow θ = nπ \pm π /3$ , $n \in Z$ .

3. Types of Trigonometric Equations

Below are the seven standard problem shapes and the technique that unlocks each.

Type 1 - Solvable by Factorisation

Bring everything to one side and factor.

Solved Example 7 (Type 1)

Solve

(2 sin x - cos x) (1 + cos x) = sin^{2} x

Solution:

Move RHS across, use $sin^{2} x = 1 - cos^{2} x = (1 - cos x) (1 + cos x)$ :

(2 sin x - cos x) (1 + cos x) - (1 - cos x) (1 + cos x) = 0

(1 + cos x) (2 sin x - 1) = 0

So $cos x = - 1$ giving $x = (2 n + 1) π$ , or $sin x = 1/2$ giving $x = nπ + (- 1)^{n} π /6$ .

Type 2 - Reducible to a Quadratic

Substitute the recurring trig ratio $t = sin x$ (or $cos x$ , etc.) and solve the resulting quadratic.

Solved Example 8 (Type 2)

Solve

2 cos^{2} x + 4 cos x = 3 sin^{2} x

Solution:

Convert $sin^{2} x = 1 - cos^{2} x$ : $2 cos^{2} x + 4 cos x - 3 + 3 cos^{2} x = 0$ , i.e. $5 cos^{2} x + 4 cos x - 3 = 0$ .

Solutions: $cos x = \frac{- 4 \pm 16 + 60}{10} = \frac{- 2 \pm 19}{5}$ .

Since $\frac{- 2 - 19}{5} \approx - 1.27 \in / [- 1, 1]$ , reject. Accept $cos x = \frac{- 2 + 19}{5}$ .

General: $x = 2 nπ \pm cos^{- 1} (\frac{- 2 + 19}{5})$ .

Type 3 - Transform Sums/Differences into Products

When you see $sin A \pm sin B$ or $cos A \pm cos B$ , use sum-to-product formulas to get a product $= 0$ , then split.

Solved Example 9 (Type 3)

Solve

cos 3 x + sin 2 x - sin 4 x = 0

Solution:

$sin 2 x - sin 4 x = - 2 cos 3 x sin x$ . So the equation becomes $cos 3 x - 2 cos 3 x sin x = 0$ , i.e. $cos 3 x (1 - 2 sin x) = 0$ .

Either $cos 3 x = 0 \Rightarrow 3 x = (2 n + 1) π /2 \Rightarrow x = (2 n + 1) π /6$ ;
or $sin x = 1/2 \Rightarrow x = nπ + (- 1)^{n} π /6$ .

Type 4 - Transform Products into Sums/Differences

Multiply both sides by $2$ where useful, then apply $2 sin A cos B = sin (A + B) + sin (A - B)$ , etc.

Solved Example 10 (Type 4)

Solve

sin 5 x cos 3 x = sin 6 x cos 2 x

Solution:

Multiply both sides by $2$ : $sin 8 x + sin 2 x = sin 8 x + sin 4 x$ , giving $sin 4 x - sin 2 x = 0$ .

Apply difference-to-product: $2 cos 3 x sin x = 0$ … wait, $sin 4 x - sin 2 x = 2 cos 3 x sin x$ . So $cos 3 x = 0$ or $sin x = 0$ .

Wait, careful: standard identity is $sin 4 x - sin 2 x = 2 cos (\frac{4 x + 2 x}{2}) sin (\frac{4 x - 2 x}{2}) = 2 cos 3 x sin x$ .

Actually simpler: $sin 4 x = 2 sin 2 x cos 2 x$ , so $sin 4 x - sin 2 x = sin 2 x (2 cos 2 x - 1)$ . Then $sin 2 x = 0$ or $cos 2 x = 1/2$ .

$sin 2 x = 0 \Rightarrow x = nπ /2$ ;
$cos 2 x = 1/2 \Rightarrow 2 x = 2 nπ \pm π /3 \Rightarrow x = nπ \pm π /6$ .

Type 5 - Equations of the Form $a sin x + b cos x = c$

Divide both sides by $a^{2} + b^{2}$ so the coefficients on the LHS become $cos φ$ and $sin φ$ ; then the LHS collapses to $sin (x + φ)$ or $cos (x - ψ)$ .

Existence: A solution exists iff

∣ c ∣ \leq a^{2} + b^{2}

. If

∣ c ∣ > a^{2} + b^{2}

the equation has no solution.

Solved Example 11 (Type 5)

Solve

sin x + cos x = 2

Solution:

Here $a^{2} + b^{2} = 2$ . Divide both sides by $2$ :

\frac{1}{2} sin x + \frac{1}{2} cos x = 1 \Rightarrow sin x cos \frac{π}{4} + cos x sin \frac{π}{4} = 1

sin (x + \frac{π}{4}) = 1 \Rightarrow x + \frac{π}{4} = (4 n + 1) \frac{π}{2} \Rightarrow x = 2 nπ + \frac{π}{4}

Solved Example 12 (Type 5)

Solve

3 cos x + 4 sin x = 5

Solution:

$a^{2} + b^{2} = 5$ , and RHS $= 5$ - the extreme value. So $cos x = 3/5$ and $sin x = 4/5$ (both must hold to reach the maximum). Setting $cos α = 3/5$ , the general solution is $x = 2 nπ + α$ where $α = tan^{- 1} (4/3)$ .

Type 6 - Polynomial in $(sin x \pm cos x)$ and $sin x cos x$

Substitute $t = sin x \pm cos x$ ; then $sin x cos x = (t^{2} - 1) /2$ (with the $+$ sign) or $(1 - t^{2}) /2$ (with $-$ ).

Solved Example 13 (Type 6)

Solve

sin x + cos x = 1 + sin x cos x

Solution:

Let $t = sin x + cos x$ . Then $t^{2} = 1 + 2 sin x cos x$ , so $sin x cos x = (t^{2} - 1) /2$ .

Equation becomes $t = 1 + (t^{2} - 1) /2 \Rightarrow 2 t = 2 + t^{2} - 1 \Rightarrow t^{2} - 2 t + 1 = 0 \Rightarrow (t - 1)^{2} = 0 \Rightarrow t = 1$ .

So $sin x + cos x = 1$ , divide by $2$ : $sin (x + π /4) = 1/ 2$ .

General: $x + π /4 = nπ + (- 1)^{n} π /4$ , giving $x = 2 nπ$ or $x = 2 nπ + π /2$ .

Type 7 - Solvable using Boundedness

When the equation forces a sum like $sin α + cos β = 2$ or similar, each term must simultaneously hit its maximum. Because $sin, cos \in [- 1, 1]$ , this gives a system that is far more constrained than either equation alone.

Solved Example 14 (Type 7)

Solve

sin 3 x + cos 2 x = - 2

Solution:

Since $sin 3 x \geq - 1$ and $cos 2 x \geq - 1$ , the sum equals $- 2$ only if both equal $- 1$ :

sin 3 x = - 1 \Rightarrow 3 x = (4 n - 1) π /2 \Rightarrow x = (4 n - 1) π /6

cos 2 x = - 1 \Rightarrow 2 x = (2 m + 1) π \Rightarrow x = (2 m + 1) π /2

Common values of $x$ : solve $(4 n - 1) /6 = (2 m + 1) /2$ , i.e. $4 n - 1 = 3 (2 m + 1) = 6 m + 3$ , i.e. $4 n = 6 m + 4$ , i.e. $2 n = 3 m + 2$ . Choose $m = 2 k$ : $2 n = 6 k + 2 \Rightarrow n = 3 k + 1$ . Substituting back, $x = (4 (3 k + 1) - 1) π /6 = (12 k + 3) π /6 = (4 k + 1) π /2$ .

General: $x = (4 k - 3) π /2$ , $k \in Z$ (equivalent forms exist).

4. Important Cautions

1. Equivalent forms exist. Different valid methods can produce different-looking general solutions. Substitute small integer values of

n

(

n = 0, \pm 1, \pm 2

) into each and check whether they generate the same set of angles in

[0, 2 π)

2. Don't cancel a common factor. Cancelling

sin x

from

tan x = 2 sin x

(giving

cos x = 1/2

) loses the solutions from

sin x = 0

. Instead, factor:

sin x (\frac{1}{cos x} - 2) = 0

, then

sin x = 0

cos x = 1/2

3. Watch domain restrictions when a factor becomes infinite. If you rewrite

sin x = 0

cos x tan x = 0

, don't blindly conclude

cos x = 0

works - at those points

tan x

is undefined, so the rewrite is not valid.

4. Squaring introduces extraneous roots. Check every candidate in the original equation. E.g.

sin θ + cos θ = 1

squared gives

sin 2 θ = 0

whose solutions include

θ = π

, but

sin π + cos π = - 1 \neq = 1

- so

π

is extraneous.

5. Respect implicit domain. If the equation contains

tan x

sec x

, take

cos x \neq = 0

; if it contains

cot x

csc x

, take

sin x \neq = 0

. If it has

lo g [f (θ)]

, require

f (θ) > 0

and base

> 0

\neq = 1

. Note

sin^{2} θ = ∣ sin θ ∣

, not

\pm sin θ

6. Verify. After finding candidates, plug back and confirm they satisfy the equation and lie in the domain.

5. Worked Practice

Solved Example 15

Solve

cos 3 x + cos^{2} x - 4 cos^{2} (x /2) = 0

Solution:

Use $2 cos^{2} (x /2) = 1 + cos x$ , so $4 cos^{2} (x /2) = 2 + 2 cos x$ . Equation:

cos 3 x + cos^{2} x - 2 - 2 cos x = 0

$cos^{2} x = (1 + cos 2 x) /2$ , so:

cos 3 x + \frac{1 + cos 2 x}{2} - 2 - 2 cos x = 0 \Rightarrow 2 cos 3 x + cos 2 x - 3 - 4 cos x = 0

Using $cos 3 x = 4 cos^{3} x - 3 cos x$ and $cos 2 x = 2 cos^{2} x - 1$ :

8 cos^{3} x - 6 cos x + 2 cos^{2} x - 1 - 3 - 4 cos x = 0 \Rightarrow 8 cos^{3} x + 2 cos^{2} x - 10 cos x - 4 = 0

Divide by $2$ : $4 cos^{3} x + cos^{2} x - 5 cos x - 2 = 0$ . Try $cos x = - 1$ : $- 4 + 1 + 5 - 2 = 0$ ✓. Factor: $(cos x + 1) (4 cos^{2} x - 3 cos x - 2) = 0$ .

$cos x = - 1 \Rightarrow x = (2 n + 1) π$ ; the quadratic gives $cos x = (3 \pm 41) /8$ ; discard the root outside $[- 1, 1]$ : $cos x = (3 - 41) /8 \approx - 0.42$ , giving $x = 2 nπ \pm cos^{- 1} ((3 - 41) /8)$ .

Solved Example 16

Solve

cot^{2} θ + 3 csc θ + 3 = 0

Solution:

Use $cot^{2} θ = csc^{2} θ - 1$ : $(cs c^{2} θ - 1) + 3 csc θ + 3 = 0 \Rightarrow csc^{2} θ + 3 csc θ + 2 = 0$ .

Factor: $(csc θ + 1) (csc θ + 2) = 0$ .

$csc θ = - 1 \Rightarrow sin θ = - 1 \Rightarrow θ = (4 n - 1) π /2 = 2 nπ - π /2$ ;
$csc θ = - 2 \Rightarrow sin θ = - 1/2 \Rightarrow θ = nπ + (- 1)^{n + 1} π /6$ .

Solved Example 17

Solve

5 sin x + 6 sin 2 x + 5 sin 3 x + sin 4 x = 0

Solution:

Group $5 (sin x + sin 3 x) + (6 sin 2 x + sin 4 x)$ :

$sin x + sin 3 x = 2 sin 2 x cos x$ ; $sin 4 x = 2 sin 2 x cos 2 x$ .

$= 10 sin 2 x cos x + 6 sin 2 x + 2 sin 2 x cos 2 x = 2 sin 2 x (5 cos x + 3 + cos 2 x)$ .

Use $cos 2 x = 2 cos^{2} x - 1$ : $= 2 sin 2 x (2 cos^{2} x + 5 cos x + 2) = 2 sin 2 x (2 cos x + 1) (cos x + 2)$ .

Since $cos x + 2 > 0$ always, discard. So $sin 2 x = 0$ or $cos x = - 1/2$ .

$sin 2 x = 0 \Rightarrow x = nπ /2$ ;
$cos x = - 1/2 \Rightarrow x = 2 nπ \pm 2 π /3$ .

Common Mistakes to Avoid

Watch out

Writing $θ = nπ + α$ for $sin θ = sin α$ . That is the formula for tangent. For sine use $θ = nπ + (- 1)^{n} α$ ; for cosine $θ = 2 nπ \pm α$ .
Dividing by a trig factor. Cancelling $sin x$ or $cos x$ loses solutions. Always factor and set each factor to zero.
Forgetting to check $∣ c ∣ \leq a^{2} + b^{2}$ in Type 5. If $c$ exceeds this bound, the equation has no solution at all.
Skipping verification after squaring. Squaring can create phantom solutions that don't satisfy the original equation.
Ignoring domain of $tan$ / $sec$ / $cot$ / $csc$ . A candidate that makes any of these undefined must be rejected.
Not simplifying $f (θ)^{2}$ correctly. $sin^{2} θ = ∣ sin θ ∣$ , not $\pm sin θ$ .

Frequently Asked Questions

What is a trigonometric equation?

A trigonometric equation is an equation involving one or more trigonometric ratios of an unknown angle. Because trig functions are periodic, such equations typically have infinitely many solutions, expressed via a general solution.

What is a principal solution?

Principal solutions are the values of the unknown that lie in the interval $[0, 2 π)$ and satisfy the equation. For example, the principal solutions of $sin x = 1/2$ are $π /6$ and $5 π /6$ .

What is the general solution?

The general solution is a formula involving an integer $n$ that gives every solution of the equation. E.g. $sin θ = sin α$ has general solution $θ = nπ + (- 1)^{n} α$ , $n \in Z$ .

What is the general solution of sin θ = sin α?

$θ = nπ + (- 1)^{n} α$ , where $α \in [- π /2, π /2]$ and $n \in Z$ . The alternating $(- 1)^{n}$ handles the two solutions per period.

What is the general solution of cos θ = cos α?

$θ = 2 nπ \pm α$ , where $α \in [0, π]$ and $n \in Z$ . The $\pm$ captures the fact that cosine is even, so both $α$ and $- α$ satisfy the equation.

What is the general solution of tan θ = tan α?

$θ = nπ + α$ , where $α \in (- π /2, π /2)$ and $n \in Z$ . Tangent has period $π$ , so a single $nπ$ shift captures every solution.

How do I solve a sin x + b cos x = c?

Divide both sides by $a^{2} + b^{2}$ to get $sin (x + φ) = c / a^{2} + b^{2}$ (where $tan φ = b / a$ ). A solution exists iff $∣ c ∣ \leq a^{2} + b^{2}$ .

Why do I get extra solutions when I square a trig equation?

Squaring can turn a false statement into a true one (e.g., $- 1 = 1$ is false, but $(- 1)^{2} = 1^{2}$ is true). Every squaring step may introduce extraneous solutions - so verify each candidate in the original equation.

Can a trigonometric equation have no solution?

Yes. For example $sin x = 2$ has no real solution because $sin x \in [- 1, 1]$ . Similarly $sin x + cos x = 3$ has none because the maximum of the LHS is $2 < 3$ .

Previous year questions on Trigonometric Equations

17 questions from past papers, each with a step-by-step solution.

Show all 17 questions

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1. Trigonometric Equation and its Solution

1.1 Principal Solutions - an Example

2. General Solutions of Standard Equations

2.1 Squared Equations

2.2 Common Special Cases

3. Types of Trigonometric Equations

Type 1 - Solvable by Factorisation

Type 2 - Reducible to a Quadratic

Type 3 - Transform Sums/Differences into Products

Type 4 - Transform Products into Sums/Differences

Type 5 - Equations of the Form asinx+bcosx=c

Type 6 - Polynomial in (sinx±cosx) and sinxcosx

Type 7 - Solvable using Boundedness

4. Important Cautions

5. Worked Practice

Common Mistakes to Avoid

Frequently Asked Questions

What is a trigonometric equation?

What is a principal solution?

What is the general solution?

What is the general solution of sin θ = sin α?

What is the general solution of cos θ = cos α?

What is the general solution of tan θ = tan α?

How do I solve a sin x + b cos x = c?

Why do I get extra solutions when I square a trig equation?

Can a trigonometric equation have no solution?

Previous year questions on Trigonometric Equations

Ready to master Trigonometry?

Type 5 - Equations of the Form $a sin x + b cos x = c$

Type 6 - Polynomial in $(sin x \pm cos x)$ and $sin x cos x$