# Solve using double angles?. Sin2x-sinx=0

Jun 18, 2018

$x = \pm 2 \pi n , \frac{\pi}{3} \pm 2 \pi n$; where $n \in \mathbb{Z}$.

#### Explanation:

We have: $\sin \left(2 x\right) - \sin \left(x\right) = 0$

Using the double-angle identity for $\sin \left(x\right)$, we get:

$R i g h t a r r o w 2 \sin \left(x\right) \cos \left(x\right) - \sin \left(x\right) = 0$

$R i g h t a r r o w \sin \left(x\right) \left(2 \cos \left(x\right) - 1\right) = 0$

$R i g h t a r r o w \sin \left(x\right) = 0$

$R i g h t a r r o w x = 0 \pm 2 \pi n$; $n \in \mathbb{Z}$

$R i g h t a r r o w x = \pm 2 \pi n$; $n \in \mathbb{Z}$

or

$R i g h t a r r o w \cos \left(x\right) = \frac{1}{2}$

$R i g h t a r r o w x = \frac{\pi}{3} \pm 2 \pi n$; $n \in \mathbb{Z}$

Therefore, the general solutions to the equation are $x = \pm 2 \pi n$ or $x = \frac{\pi}{3} \pm 2 \pi n$, where $n \in \mathbb{Z}$.

Jun 18, 2018

$x = k \pi \text{ or } x = \frac{\pi}{3} + 2 k \pi$

#### Explanation:

$\text{using the "color(blue)"trigonometric identity}$

â€¢color(white)(x)sin2x=2sinxcosx

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

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

$\text{equate each factor to zero and solve for } x$

$\sin x = 0 \Rightarrow x = k \pi \to \left(k \in \mathbb{Z}\right)$

$\cos x = \frac{1}{2} \Rightarrow x = \frac{\pi}{3} + 2 k \pi \to \left(k \in \mathbb{Z}\right)$