General Solution of sin x =0, cos x=0, tan x=0

One can define equations involving trigonometric functions. These equations are called trigonometric equations. In this post, we will learn about the general solutions of the trigonometric equations: sin x=0, cos x =0, and tan x =0.

Solution of sin x =0

Solve $\sin x=0$.

The general solutions of the equation $\sin x=0$ are the integer multiples of $\pi$. In other words:

$\sin x=0$ if and only if $x=n\pi$ where n is an integer.

So the solutions of sin x =0 are x=nπ where n is an integer.

Solution of cos x =0

Solve $\cos x=0$.

The general solutions of the equation $\cos x=0$ are the odd multiples of ${\displaystyle \frac{\pi }{2}}$. In other words:

$\cos x=0$ if and only if $x=(2n+1){\displaystyle \frac{\pi }{2}}$ where some integer n.

Thus the solutions of cos x =0 are x=(2n+1)${\displaystyle \frac{\pi }{2}}$ where n is an integer.

Solution of tan x =0

Solve tan x =0.

Note that $\tan x=0$

$\Rightarrow {\displaystyle \frac{\sin x}{\cos x}}=0$

$\Rightarrow \sin x=0$

Therefore, the general solutions of the equation $\tan x=0$ are given by the solutions of the equation $\sin x=0$.

From above we know that $\sin x=0$  $⟺x=n\pi$ where $n\in Z$. So the general solutions of $\tan x=0$ are the integer multiples of $\pi$. In other words:

$\tan x=0$ if and only if $x=n\pi$ where n is an integer.

Hence the solutions of tan x =0 are x=n$\pi$ where n is an integer.

Example1: Solve sin 2x =0.

Solution:

As the solutions of sin x =0 are x=n$\pi$, we deduce that the solutions of sin 2x =0 are

$2x=n\pi$

$\Rightarrow x={\displaystyle \frac{n\pi }{2}}$

So the solutions of sin 2x =0 are $x={\displaystyle \frac{n\pi }{2}}$ where n is an integer.

Example2: Solve sin 3x =0.

Solution:

We know that the solutions of sin x =0 are x=n$\pi$. Thus the solutions of sin 3x =0 are

$3x=n\pi$

$\Rightarrow x={\displaystyle \frac{n\pi }{3}}$

Thus the solutions of sin 3x =0 are $x={\displaystyle \frac{n\pi }{3}}$ where n is an integer.

Example3: Solve cos 3x =0.

Solution:

We know that the solutions of cos x =0 are x=(2n+1)${\displaystyle \frac{\pi }{2}}$. Thus the solutions of cos 3x =0 are

$3x=(2n+1){\displaystyle \frac{\pi }{2}}$

$\Rightarrow x=(2n+1){\displaystyle \frac{\pi }{6}}$

So the solutions of cos 3x =0 are $x=(2n+1){\displaystyle \frac{\pi }{6}}$ where n is an integer.

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