# Solve the given equation. If there is no solution, enter NO SOLUTION: cos2x=5sinx-2?

## I don't know how to start the problem.

Dec 3, 2017

Answer: $x = \frac{\pi}{6} + 2 \pi k , \frac{5 \pi}{6} + 2 \pi k$, $k \in \mathbb{Z}$

#### Explanation:

Solve $\cos 2 x = 5 \sin x - 2$

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

Substituting in, we have:
$1 - 2 {\sin}^{2} \left(x\right) = 5 \sin x - 2$

Moving everything to one side:
$2 {\sin}^{2} \left(x\right) + 5 \sin x - 3 = 0$

We now have a quadratic which we can factor:
$\left(2 \sin x - 1\right) \left(\sin x + 3\right) = 0$

So, we have that $\sin x = \frac{1}{2}$ or $\sin x = - 3$, clearly the latter is not possible since $\sin x \in \left[- 1 , 1\right]$

Therefore $x = \frac{\pi}{6} + 2 \pi k , \frac{5 \pi}{6} + 2 \pi k$, $k \in \mathbb{Z}$