Question

Solve the equation
$\left(v\right)\sin x-3\sin 2x+\sin 3x=\cos x-3\cos 2x+\cos 3x$

Answer 1

$\sin x-3\sin 2x+\sin 3x=\cos x-3\cos 2x+\cos 3x$

$\Rightarrow \sin x+\sin 3x-3\sin 2x=\cos x+\cos 3x-3\cos 2x$

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

$\Rightarrow 2\sin \left(\frac{3x+x}{2}\right)\cos \left(\frac{3x-x}{2}\right)-3\sin 2x=2\cos \left(\frac{3x+x}{2}\right)\cos \left(\frac{3x-x}{2}\right)-3\cos 2x$

$\Rightarrow 2\sin 2x\cos x-3\sin 2x=2\cos 2x\cos x-3\cos 2x$

$\Rightarrow \sin 2x(2\cos x-3)=\cos 2x(2\cos x-3)$

$\Rightarrow \sin 2x(2\cos x-3)-\cos 2x(2\cos x-3)=0$

$\Rightarrow (\sin 2x-\cos 2x)(2\cos x-3)=0$

$\Rightarrow (\sin 2x-\cos 2x)=0\text{or}(2\cos x-3)=0$

$\Rightarrow \tan 2x=1\text{or}\cos x=\frac{3}{2}$

As $\cos x\in [-1,1]$, so $\tan 2x=1$

$\Rightarrow \tan 2x=\tan \frac{\pi }{4}$

We know the general solution of $\tan x=\tan \alpha$ is $x=n\pi +\alpha ,n\in Z$

So,
$2x=n\pi +\frac{\pi }{4},n\in Z$

$\Rightarrow x=\frac{n\pi }{2}+\frac{\pi }{8}$

Hence, the general solution is
$x=\frac{n\pi }{2}+\frac{\pi }{8},n\in Z$