An equation involving one or more trigonometrical ratios of unknown angle is called a **trigonometric equation **

e.g. cos^{2}x – 4 sinx = 1

It is to be noted that a trigonometrical identity is satisfied for every value of the unknown angle whereas trigonometric equation is satisfied only for some values (finite or infinite) of unknown angle.

e.g. sin^{2}x + cos^{2}x = 1 is a trigonometrical identity as it is satisfied for every value of x ∈ R.* *

*Solution of a Trigonometric Equation:*

A value of the unknown angle which satisfies the given equation is called a solution of the equation

e.g. sinθ = 1/2 ⇒ θ = π/6.** **

*General Solution:*

Since trigonometrical functions are periodic functions, solutions of trigonometric equations can be generalized with the help of the periodicity of the trigonometrical functions.

The solution consisting of all possible solutions of a trigonometric equation is called its **general solution.**

We use the following formulae for solving the trigonometric equations:

sinθ = 0 ⇒ θ = nπ

cosθ = 0 ⇒ θ = (2n+1)π/2

tanθ = 0 ⇒ θ = nπ

sinθ = sinα ⇒ θ = nπ + (– 1)^{n}α , where α ∈ [– π/2 , π/2]

cosθ = cosα ⇒ θ = 2nπ ± α , where α ∈ [ 0 , π]

tanθ = tanα ⇒ θ = nπ + α , where α ∈ (– π/2, π/2)

sin^{2}θ = sin^{2}α , cos^{2}θ = cos^{2}α , tan^{2} θ = tan^{2}α ⇒ θ = nπ ± α ,

sinθ = 1 ⇒ θ = (4n + 1)π/2

sinθ = –1 ⇒ θ = (4n – 1)π/2

cosθ = 1 ⇒ θ = 2nπ

cosθ = – 1 ⇒ θ = (2n + 1)π

sinθ = sinα and cosθ = cosα ⇒ θ = 2nπ + α

### Note:

Everywhere in this chapter n is taken as an integer, If not stated otherwise.

The general solution should be given unless the solution is required in a specified interval or range.

α is taken as the principal value of the angle. Numerically least angle is called the principal value.

### Important points to Remember:

* While solving a trigonometric equation, squaring the equation at any step should be avoided as far as possible. If squaring is necessary, check the solution for extraneous values.

*Never cancel terms containing unknown terms on the two sides, which are in product. It may cause loss of genuine solution.

*The answer should not contain such values of angles which make any of the terms

*Domain should not be changed. If it is changed, necessary corrections must be incorporated.

*Check that denominator is not zero at any stage while solving equations.

*At times you may find that your answers differ from those in the package in their notations. This may be due to the different methods of solving the same problem. Whenever you come across such situation, you must check their authenticity. This will ensure that your answer is correct.

*While solving trigonometric equations, you may get same set of solution repeated in your answer. It is necessary for you to exclude these repetitions. e.g. in nπ + π/2 , kπ/5 + π/10 (n , k ∈ I ) , the set nπ + π/2 forms a part of the second set of solution ( you can check by putting k = 5m +2 (m ∈ I). Hence the final answer should be kπ/5 + π/10 , k ∈ I

* Sometimes the two solution set consist partly of common values. In all such cases the common part must be presented only once.

**Solving the different forms of trigonometric equations :**

**Equations Reducible to Lower Degree:**

**Illustration : **Solve the trigonometric equation

$\large sin^4 x + sin^4 (x+\frac{\pi}{4}) = \frac{1}{4} $

**Solution: **Given trigonometric equation can be written as

$\large 4 sin^4 x + 4 sin^4 (x+\frac{\pi}{4}) = 1 $

$\large (1-cos2x)^2 + (1- cos(2x+\frac{\pi}{2}))^2 = 1$

⇒ (1 − cos2x)^{2} + (1 + sin2x)^{2} = 1

⇒ 2 + 1 − 2(cos2x − sin2x) = 1

⇒ cos(2x + π/4) = 1/√2

⇒ 2x + π/4 = 2nπ ± π/4

∴ x = nπ , nπ − π/4

**Equations Reducible by Direct Formula / Multiple Angle Formula:**

**Illustration : *** Solve the trigonometric equation *

$\large cot\frac{\theta}{2} – cosec\frac{\theta}{2} = cot\theta$

**Solution :**

$\large cot\frac{\theta}{2} – cot\theta = cosec\frac{\theta}{2} $

$\large \frac{cos\theta/2}{sin\theta/2} – \frac{cos\theta}{sin\theta} = cosec\frac{\theta}{2} $

$\large \frac{cos\theta/2}{sin\theta/2} – \frac{cos\theta}{2sin\theta/2 . cos\theta/2} = cosec\frac{\theta}{2} $

$\large \frac{2cos^2 (\theta/2) – cos\theta}{sin\theta} = cosec\frac{\theta}{2} $

$\large \frac{2cos^2 (\theta/2) – cos\theta}{sin\theta} = \frac{1}{sin(\theta/2)} $

$\large 2cos^2 (\theta/2) sin(\theta/2) – cos\theta sin(\theta/2) = sin\theta $

$\large 2cos^2 (\theta/2) sin(\theta/2) – cos\theta sin(\theta/2) = 2 sin(\theta/2) cos(\theta/2) $

$\large sin(\theta/2) [ 2cos^2 (\theta/2) -( 2cos^2(\theta/2)-1 ) ] = 2 sin(\theta/2) cos(\theta/2) $

$\large sin(\theta/2) = 2 sin(\theta/2) cos(\theta/2) $

$\large 2 sin(\theta/2) cos(\theta/2) -sin(\theta/2) = 0 $

$\large sin(\theta/2) [cos(\theta/2) -1 ] = 0 $

sin(θ/2) = 0 ⇒ θ = 2 n π

cos(θ/2 )= 1/2 ⇒ θ /2 = 2 n π ± π/3

θ = 4 n π ± 2π/3

for θ = 2 n π ; cot (θ/2) is undefined. Hence do not satisfy the domain of given equation.

The only solution is θ = 4 n π ± 2π/3 , where n ∈ I.