Question

Solve the equation
$\left(ix\right)\sin 2x-\sin 4x+\sin 6x=0$

Answer 1

$\sin 2x-\sin 4x+\sin 6x=0$

$\Rightarrow (\sin 6x+\sin 2x)-\sin 4x=0$

Using $\sin C+\sin D=2\sin \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)$

$\Rightarrow 2\sin \left(\frac{6x+2x}{2}\right)\cos \left(\frac{6x-2x}{2}\right)-\sin 4x=0$

$\Rightarrow 2\sin 4x\cos 2x-\sin 4x=0$

$\Rightarrow \sin 4x(2\cos 2x-1)=0$

$\Rightarrow \sin 4x=0\text{or}(2\cos 2x-1)=0$

$\Rightarrow 4x=n\pi \text{or}\cos 2x=\frac{1}{2}$

$\Rightarrow x=\frac{n\pi }{4}\text{or}\cos 2x=\cos \frac{\pi }{3}$

We know the general solution of $\cos x=\cos \alpha$ is $x=2n\pi \pm \alpha ,m\in Z$

So, the general solution of $\cos 2x=\cos \frac{\pi }{3}$ is
$2x=2m\pi \pm \frac{\pi }{3}\Rightarrow x=m\pi \pm \frac{\pi }{6}$

Hence, the general solution is
$x=\frac{n\pi }{4}\text{and}x=m\pi \pm \frac{\pi }{6},n,m\in Z$