Trigonometric Equations

An equation involving trigonometric functions of a variable is called a trigonometric equation.
A solution of trigonometric equation is a value of variable that satisfies the equation.
Since all trigonometric functions are periodic, usually the trigonometric equations have unlimited number of  solutions.
The solutions lying between 0 to 2π (0≤ x ≤2π) are called principal solutions.
A solution generalized by means of periodicity, involving integer 'n', is called a general solution.

Solving sin x = 0, cos x = 0 and tan x = 0
We know that,
sin x = 0, when x = 0, ±π, ±2π, ±3π,...
Thus, sin x=0 ⇒ x = nπ, n∈I

cos x=0 when $x=\pm \frac{\pi }{2},\pm \frac{3\pi }{2},\pm \frac{5\pi }{2},...$
Thus, cos x=0 ⇒  $x=\left ( 2n+1 \right )\frac{\pi }{2}$, n∈I

tan x=0 when x = 0, ±π, ±2π, ±3π,...
Thus, tan x=0 ⇒  x = nπ, n∈I

For all real numbers x and y
sin x = sin y implies $x=n\pi +(-1)^{n}y$
Proof: Given, sin x = sin y
⇒ $\sin x-\sin y=0$
⇒ $2\cos (\frac{x+y}{2})\sin (\frac{x-y}{2})=0$
⇒ $\cos (\frac{x+y}{2})=0$  or $\sin (\frac{x-y}{2})=0$
⇒ $\frac{x+y}{2}=(2n+1)\frac{\pi}{2}$  or $\frac{x-y}{2}=n\pi$ [cos x=0 ⇒  $x=\left ( 2n+1 \right )\frac{\pi }{2}$ & sin x=0 ⇒ x = nπ]
⇒ $x+y=(2n+1)\pi$  or $x-y=2n\pi$
⇒ $x=(2n+1)\pi-y$  or $x=2n\pi+y$
⇒ $x=(2n+1)\pi+(-1)^{2n+1}y$ or $x=2n\pi+(-1)^{2n}y$
Combining both results, we get,
$x=n\pi+(-1)^{n}y$ 

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cos x = cos y implies $x=2n\pi\pm y$
Proof: Given, cos x = cos y
⇒ $\cos x-\cos y=0$
⇒ $-2\sin (\frac{x+y}{2})\sin (\frac{x-y}{2})=0$
⇒ $\sin (\frac{x+y}{2})=0$  or $\sin (\frac{x-y}{2})=0$
⇒ $\frac{x+y}{2}=n\pi$  or $\frac{x-y}{2}=n\pi$ [sin x=0 ⇒ x = nπ]
⇒ $x+y=2n\pi$  or $x-y=2n\pi$
⇒ $x=2n\pi-y$  or $x=2n\pi+y$
Combining both results, we get,
$x=2n\pi\pm y$

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tan x = tan y implies $x=n\pi+y$ 
Proof: Given, tan x = tan y
⇒ $\frac{\sin x}{\cos x}=\frac{\sin y}{\cos y}$
⇒ $\sin x \cos y=\sin y \cos x$
⇒ $\sin x \cos y- \cos x \sin y=0$
⇒ $\sin (x-y)=0$ [sin(x-y)=sin x cos y- cos x sin y]
⇒  $x-y=n\pi$
$x=n\pi+y$