# How do you solve sin 4beta = cos 5beta?

Aug 18, 2016

$\frac{\pi}{18} + 2 k \pi$
$\frac{3 \pi}{2} + 2 k \pi$

#### Explanation:

sin 4x = cos 5x (1)
Use trig identity:
$\sin a = \cos \left(\frac{\pi}{2} - a\right)$
$\sin 4 x = \cos \left(\frac{\pi}{2} - 4 x\right)$
From (1):
$\cos 5 x = \cos \left(\frac{\pi}{2} - 4 x\right)$
$5 x = \pm \left(\frac{\pi}{2} - 4 x\right)$
a. $5 x = \frac{\pi}{2} - 4 x$--> $9 x = \frac{\pi}{2}$ --> $x = \frac{\pi}{18}$
b. $5 x = - \frac{\pi}{2} + 4 x$ --> $x = - \frac{\pi}{2}$, or $x = \frac{3 \pi}{2}$
$x = \frac{\pi}{18} + 2 k \pi$
$x = \frac{3 \pi}{2} + 2 k \pi$
$x = \frac{\pi}{18} = {10}^{\circ}$ --> sin 4x = sin 40 = 0.642