# How do you solve sin(x)-cos(x/3)=0?

May 17, 2018

$x = \frac{3 \pi}{8}$

#### Explanation:

.

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

$\sin x = \cos \left(\frac{x}{3}\right)$

We know:

$\sin \left(\frac{\pi}{2} - \theta\right) = \cos \theta$

Therefore,

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

$x = \frac{\pi}{2} - \frac{x}{3}$

$x + \frac{x}{3} = \frac{\pi}{2}$

$6 x + 2 x = 3 \pi$

$8 x = 3 \pi$

$x = \frac{3 \pi}{8}$

May 17, 2018

$x = \frac{3 \pi}{8} + \frac{3 k \pi}{2}$
$x = \frac{3 \pi}{4} + 3 k \pi$

#### Explanation:

$\sin x = \cos \left(\frac{x}{3}\right)$
$\cos \left(\frac{\pi}{2} - x\right) = \cos \left(\frac{x}{3}\right)$
Unit circle and property of cos function -->
$\frac{\pi}{2} - x = \pm \frac{x}{3}$
a. $\frac{\pi}{2} - x = \frac{x}{3}$
$x + \frac{x}{3} = \frac{\pi}{2}$
$\frac{4 x}{3} = \frac{\pi}{2} + 2 k \pi$
$x = \frac{3 \pi}{8} + \frac{3 k \pi}{2}$
b. $\frac{\pi}{2} - x = - \frac{x}{3}$
$x - \frac{x}{3} = \frac{\pi}{2}$
$\left(2 \frac{x}{3}\right) = \frac{\pi}{2} + 2 k \pi$
$x = \frac{3 \pi}{4} + 3 k \pi$