Question

Solve the equation
$\left(vi\right)\sin x+\sin 2x+\sin 3x=0$

Answer 1

$\sin x+\sin 2x+\sin 3x=0$

$\Rightarrow (\sin 3x+\sin x)+\sin 2x=0$

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

$\Rightarrow 2\sin \left(\frac{3x+x}{2}\right)\cos \left(\frac{3x-x}{2}\right)+\sin 2x=0$

$\Rightarrow 2\sin 2x\cos x+\sin 2x=0$

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

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

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

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

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

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

Hence, the general solution is
$x=\frac{n\pi }{2}\text{and}x=2m\pi \pm \frac{2\pi }{3},n,m\in Z$