Question

prove that sin7x=sin3x

Answer 1

tell me if you find any problem here.

Answer 2

Hence may be value of x be in the form of  $x= \frac{n\pi }{2}$ or

in the form of $x=\frac{2n\pi}{5} \± \frac{\pi}{10}$.

Step-by-step explanation:

Given:

We have,

sin7x=sin3x

⇒ sin7x-sin3x=0

⇒ $2[cos(\frac{7x+3x}{2} )sin(\frac{7x-3x}{2} )]=0$   $[sinC-sinD=2cos(\frac{C+D}{2} )sin(\frac{C-D}{2} )]$

⇒ $2[cos(\frac{7x+3x}{2} )sin(\frac{7x-3x}{2} )]=0$

⇒ $2[cos(\frac{10x}{2} )sin(\frac{4x}{2} )]=0$

⇒ 2 cos5x sin2x = 0

⇒ $cos5x sin2x = \frac{0}{2}$

⇒ cos5x sin2x = 0

Therefore, cos5x = 0 or sin2x = 0

So, we know that whenever the value of angle equals to nπ then sinx is equal to zero.

⇒ 2x = nπ

⇒ $x= \frac{n\pi }{2}$

Also, we know that whenever the value of angle equals to $(n \pi\±\frac{\pi}{2})$ then cos x is equal to zero.

$5x = (2n \pi\±\frac{\pi}{2})$

⇒ $x=\frac{2n\pi}{5} \± \frac{\pi}{10}$

Hence may be value of x be in the form of  $x= \frac{n\pi }{2}$ or

in the form of $x=\frac{2n\pi}{5} \± \frac{\pi}{10}$.