# How do you solve Sin(3x) - sin(6x) = 0?

May 14, 2016

$0 , \frac{\pi}{3} , \frac{2 \pi}{3} , \frac{\pi}{9} , \mathmr{and} \frac{5 \pi}{9}$

#### Explanation:

Use the trig identity: sin 2a + 2sin a.cos a.
Replace in the equation sin (6x) by 2sin (3x).cos (3x) -->
sin (3x) - 2sin (3x).cos (3x) = 0
sin (3x)(1 - 2cos 3x) = 0
a. sin 3x = 0 --> 3x = 0 and $3 x = \pi$, and $3 x = 2 \pi$, that give as answers-->
x = 0; x = pi/3, and x = (2pi)/3
b. $\cos \left(3 x\right) = \frac{1}{2}$ --> $3 x = \pm \frac{\pi}{3}$ -->
$x = \frac{\frac{\pi}{3}}{3} = \frac{\pi}{9}$ and $x = \frac{\frac{5 \pi}{3}}{3} = \frac{5 \pi}{9}$
Note: the arc $- \left(\frac{\pi}{3}\right)$ is co-terminal to the arc $\frac{5 \pi}{3.}$