Question

sin5x+sin2x+sin3x+sin4x=0

Answer 1

$\sin5x+\sin2x+\sin3x+\sin4x=0$
$(\sin5x+\sin3x)+(\sin2x+\sin4x)=0$
$2\sin\frac{5x+3x}{2} \cos\frac{5x-3x}{2} +2\sin\frac{2x+4x}{2} \cos\frac{2x-4x}{2} =0$
$\sin4x \cos x+\sin3x \cos (-x)=0$
$\sin4x \cos x+\sin3x \cos x=0$
$\cos x(\sin4x +\sin3x )=0$
$\left[\begin{array}{l} \cos x=0\\ \sin4x +\sin3x=0 \end{array}$
$\left[\begin{array}{l} \cos x=0\\ 2\sin \frac{4x+3x}{2} \cos \frac{4x-3x}{2} =0 \end{array}$
$\left[\begin{array}{l} \cos x=0 \\ \sin \frac{7x}{2}=0 \\ \cos \frac{x}{2} =0 \end{array}$
$\left[\begin{array}{l} x= \frac{ \pi }{2}+\pi n \\\\ \frac{7x}{2}= \pi n \\\\ \frac{x}{2} = \frac{ \pi }{2}+ \pi n \end{array} \Rightarrow \left[\begin{array}{l} x_1= \frac{ \pi }{2}+\pi n \\\\ x_2= \frac{ 2\pi n}{7} \\\\ x_3= \pi + 2\pi n \end{array}, n\in Z$