Question

Find the general solution of the equation $\sin x+\sin 3x+\sin 5x=0$

Answer 1

Step 1: Simplication
Given : $\sin x+\sin 3x+\sin 5x=0$
$\Rightarrow (\sin x+\sin 5x)+\sin 3x=0$
$\Rightarrow 2\sin \left(\frac{5x+x}{2}\right)\cos \left(\frac{x-5x}{2}\right)+\sin 3x=0$
$\Rightarrow 2\sin \left(\frac{6x}{2}\right)\cos \left(\frac{-4x}{2}\right)+\sin 3x=0$
$\Rightarrow 2\sin 3x\cos 2x+\sin 3x=0$
$\Rightarrow \sin 3x(2\cos 2x+1)=0$
$\sin 3x=0$ or $2\cos 2x+1=0$
$\sin 3x=0$ or $\cos 2x=-\frac{1}{2}$

Step $2:$
General solution for $\sin 3x=0$
General solution is $3x=n\pi ,n\in Z$
$x=\frac{n\pi }{3},$ where $n\in Z$

Step $3:$
General solution for $\cos 2x=-\frac{1}{2}$
$\cos 2x=-\frac{1}{2}$
$\Rightarrow \cos 2x=\cos \left(\frac{2\pi }{3}\right)$

We know that general solution for $\cos x=\cos y$ is $x=2n\pi \pm y,n\in Z$
So, the general solution is $2x=2n\pi \pm \frac{2\pi }{3},n\in Z$
$x=n\pi \pm \frac{\pi }{3}$ where $n\in Z$