# Question #bb3f7

Jan 17, 2016

$x = \frac{\pi}{2} , \pi$

#### Explanation:

The first step is to rewrite $\cos \left(2 x\right)$ in terms of $\sin \left(x\right)$.

• $\cos \left(2 x\right) = {\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right)$
• ${\cos}^{2} \left(x\right) = 1 - {\sin}^{2} \left(x\right)$
• $\cos \left(2 x\right) = 1 - {\sin}^{2} \left(x\right) - {\sin}^{2} \left(x\right) = 1 - 2 {\sin}^{2} \left(x\right)$

Thus, the function can be rewritten as

$1 - 2 {\sin}^{2} \left(x\right) + 2 \sin \left(x\right) - 3 = - 2$

Move all the terms to the same side.

$0 = 2 {\sin}^{2} \left(x\right) - 2 \sin \left(x\right)$

Divide both sides by $2$.

${\sin}^{2} \left(x\right) - \sin \left(x\right) = 0$

Factor a $\sin \left(x\right)$ term.

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

Here, we have a product of two terms that equals $0$. This means that either one of the two terms could be equal to $0$:

$\sin x = 0 \textcolor{w h i t e}{s s s s s s} \text{or} \textcolor{w h i t e}{s s s s s s} \sin \left(x\right) - 1 = 0 \implies \sin \left(x\right) = 1$
$x = \pi \textcolor{w h i t e}{s s s s s s s k l} \text{or} \textcolor{w h i t e}{s s s s s s s s s s s s s s s s s s s s s s s s} x = \frac{\pi}{2}$

The times when $\sin \left(x\right) = 0 , 1$ for $0 < x < 2 \pi$ are $x = \frac{\pi}{2} , \pi$.