# The general solution of equation sinx+sin5x=sin2x+sin4x is?

Jun 22, 2018

f(x) = sin x + 5x = sin 2x + sin 4x
Use trig identity:
sin a + sin b = 2sin ((a + b)/2)cos ((a - b)/2
Left side:
sin x + sin 5x = 2sin 3x.cos 2x
Right side:
sin 2x + sin 4x = 2sin 3x.cos x
f(x) = 2sin 3x.cos 2x - 2sin 3x.cos x = 0
(2sin 3x)(cos 2x - cos x) = 0
Either factor should be zero.
a. sin 3x = 0
$x = 2 k \pi$, and $x = \pi + 2 k \pi = \left(2 k + 1\right) \pi$, and $x = 2 k \pi$
b. cos 2x - cos x = 0
Reminder of trig identity:
$\cos a - \cos b = - 2 \sin \left(\frac{a + b}{2}\right) . \sin \left(\frac{a - b}{2}\right)$
In this case:
$\cos 2 x - \cos x = - 2 \sin \left(3 \frac{x}{2}\right) . \sin \left(\frac{x}{2}\right)$
a. $\sin \left(3 \frac{x}{2}\right) = 0$ -->
$\left(3 \frac{x}{2}\right) = 2 k \pi$ --> $x = \frac{4 k \pi}{3}$
$\left(3 \frac{x}{2}\right) = \pi + 2 k \pi = \left(2 k + 1\right) \pi$ --> $x = \left(2 k + 1\right) \left(2 \frac{\pi}{3}\right)$
b. $\sin \left(\frac{x}{2}\right) = 0$
$\frac{x}{2} = 2 k \pi$--> $x = 4 k \pi = 2 n \pi$
$\frac{x}{2} = \left(2 k + 1\right) \pi$ --> $x = \left(2 k + 1\right) 2 \pi = 2 n \pi$