# Trigonometric Equations

The equations that involve the trigonometric functions of a variable are called trigonometric equations. In the upcoming discussion, we will try to find the solutions of such trigonometric equations. We know that sin x and cos x repeat themselves after an interval of 2π and tan x repeats itself after an interval of π.

The solutions of trigonometric equations which lie in the interval of [0, 2π] are called principal solutions. A trigonometric equation will also have a general solution expressing all the values which would satisfy the given trigonometric equation and it is expressed in a generalized form in terms of ‘n’.

Let us begin with a basic equation, sin x = 0.The principal solution for this case will be x = 0,π,2π as these values satisfy the given equation lying in the interval [0, 2π] . But, we know that if sin x = 0, then x = 0, π, 2π, π, -2π, -6π, etc. are solutions of the given equation. Hence general solution for sin x = 0 will be, x = nπ, where n∈I.

Similarly, general solution for cos x = 0 will be x = (2n+1)π/2, n∈I, as cos x has a value equal to 0 at π/2, 3π/2, 5π/2, -7π/2, -11π/2 etc.

Now, let us consider the equation, sin x = sin y. Let us try to find the general solution for this trigonometric equation.

sin x = sin y

⇒ sin x – sin y = 0

⇒2cos (x + y)/2 sin (x – y)/2 = 0

⇒cos (x + y)/2 = 0 or sin (x – y)/2 = 0

Upon taking the common solution from both the conditions, we get:

x = nπ + (-1)^{n}y, where n∈I

Similarly, the general solution of cos x = cos y will be:

cos x – cos y = 0

2sin (x + y)/2 sin (y – x)/2 = 0

sin (x + y)/2 = 0 or sin (x – y)/2 = 0

(x + y)/2 = nπ or (x – y)/2 = nπ

On taking the common solution from both the conditions, we get:

x = 2nπ± y, where n∈I

Similarly to find the solution of equations involving tan x or other functions, we can use conversion of trigonometric equations. In other words, if tan x = tan y then;

Let us go through an example to have a better insight about solutions of trigonometric equations.

**Example:** sin 2x – sin 4x + sin 6x = 0

**Solution: **Given: sin 2x – sin 4x + sin 6x = 0

⇒sin 2x – sin 4x + sin 6x = 0

⇒2sin 4x.cos 2x – sin 4x = 0

⇒sin 4x (2cos 2x – 1) = 0

⇒sin 4x = 0 or cos 2x = ½

⇒4x = nπ or 2x = 2nπ ± π/3

Therefore the general solution for the given trigonometric equation is:

⇒x = nπ/4 or nπ ± π/6