Question

Solve the equation
$\left(vii\right)\sin x+\sin 2x+\sin 3x+\sin 4x=0$

Answer 1

$\sin x+\sin 2x+\sin 3x+\sin 4x=0$

$\Rightarrow (\sin 3x+\sin x)+(\sin 4x+\sin 2x)=0$

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

$\Rightarrow 2\sin \left(\frac{3x+x}{2}\right)\cos \left(\frac{3x-x}{2}\right)+2\sin \left(\frac{4x+2x}{2}\right)\cos \left(\frac{4x-2x}{2}\right)=0$

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

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

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

$\Rightarrow \cos x\left[\sin \left(\frac{5x}{2}\right)\cos \left(\frac{x}{2}\right)\right]=0$

$\Rightarrow \cos x\sin \left(\frac{5x}{2}\right)\cos \left(\frac{x}{2}\right)=0$

$\Rightarrow \cos x=0\text{or}\sin \left(\frac{5x}{2}\right)=0\text{or}\cos \left(\frac{x}{2}\right)=0$

$\Rightarrow x=(2n+1)\frac{\pi }{2}\text{or}\frac{5x}{2}=m\pi \text{or}\frac{x}{2}=(2p+1)\frac{\pi }{2}$

Hence, the general solution is
$x=(2n+1)\frac{\pi }{3}\text{or}x=\frac{2m\pi }{5}\text{or}x=(2p+1)\pi \text{Where},n,m,p\in Z$