Question

Find the General solution of
sin2x + sin4x + sin6x = 0 ​

Answer 1

Step-by-step explanation:

your answer attached in the photo

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

We can also write it as :-

(sin6x + sin2x) + sin4x = 0

Formula :-

${\boxed{\sf\:{sinC+sinD=2sin\dfrac{(C+D)}{2}cos\dfrac{(C-D)}{2}}}}$

So,

$\tt{\rightarrow 2sin\dfrac{6x+2x}{2}cos\dfrac{6x-2x}{2}+sin4x=0}$

2sin4xcos2x + sin4x = 0

sin4x(2cos2x + 1) = 0

sin4x = 0

2cos2x + 1 = 0

$\tt{\rightarrow cos2x=-\dfrac{1}{2}}$

$\tt{\rightarrow cos2x=-cos\dfrac{\pi}{3}}$

$\tt{\rightarrow cos2x=cos(\pi+\dfrac{\pi}{3})}$

$\tt{\rightarrow cos2x=cos\dfrac{2\pi}{3}}$

Also,

sin4x = 0

Or,

$\tt{\rightarrow cos2x=cos\dfrac{2\pi}{3}}$

4x = nπ

Or,

$\tt{\rightarrow 2x=(2m\pi \pm\dfrac{2\pi}{3})}$

Also,

$\tt{\rightarrow x=\dfrac{n\pi}{4}}$

$\tt{\rightarrow x=(m\pi \pm\dfrac{\pi}{3})}$

${\boxed{\bigstar{{General\;Solution :- }}}}$

$\Large{\boxed{\sf\:{x=\dfrac{n\pi}{4}}}}$

$\Large{\boxed{\sf\:{x=(m\pi \pm\dfrac{\pi}{3}}}}$