# How do you solve the following equation sin 2x - sin x = 0 in the interval [0, 2pi]?

Jun 20, 2016

Use the identity $\sin 2 \theta = 2 \sin \theta \cos \theta$.

#### Explanation:

$2 \sin x \cos x - \sin x = 0$

$\sin x \left(2 \cos x - 1\right) = 0$

$\sin x = 0 \mathmr{and} \cos x = \frac{1}{2}$

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

Hopefully this helps!

Jun 20, 2016

The soln. set $= \left\{0 , \frac{\pi}{3} , \pi , 5 \frac{\pi}{3} , 2 \pi\right\} .$

#### Explanation:

Given that, $\sin 2 x - \sin x = 0. \Rightarrow 2 \sin x \cos x - \sin x = 0. \Rightarrow \sin x \left(2 \cos x - 1\right) = 0. \Rightarrow \sin x = 0 , \mathmr{and} , \cos x = \frac{1}{2}$

The zeros of sin are, $k \pi , k \in \mathbb{Z} .$ Since we require zeros in $\left[0 , 2 \pi\right]$, these are $0 , \pi , 2 \pi .$

Next $\cos x = \frac{1}{2} = \cos \left(\frac{\pi}{3}\right) \Rightarrow x = 2 k \pi \pm \frac{\pi}{3} , k \in \mathbb{Z} .$This is since that the general soln. of the eqn. $\cos \theta = \cos \alpha$ is $\theta = 2 k \pi \pm \alpha , k \in \mathbb{Z}$.

$k = 0 , \Rightarrow x = + \frac{\pi}{3} \in \left[0 , 2 \pi\right] ,$ as $- \frac{\pi}{3} \notin \left[0 , 2 \pi\right]$
$k = 1 \Rightarrow x = 2 \pi - \frac{\pi}{3} = 5 \frac{\pi}{3} ,$ as $2 \pi + \frac{\pi}{3} \notin \left[0 , 2 \pi\right] .$

Altogether, the soln. set $= \left\{0 , \frac{\pi}{3} , \pi , 5 \frac{\pi}{3} , 2 \pi\right\} .$