# How do you find the critical numbers of y=cos x + sin 2x?

##### 1 Answer
Jan 13, 2018

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

$y ' = - 4 {\sin}^{2} x - \sin x + 2$

$y ' = 0$ at solutions to

$4 {\sin}^{2} x + \sin x - 2 = 0$

$\sin x = \frac{- 1 \pm \sqrt{1 + 32}}{8} = \frac{- 1 \pm \sqrt{33}}{8}$

$x = \left\{\begin{matrix}{\sin}^{-} 1 \frac{- 1 - \sqrt{33}}{8} + 2 \pi k \\ - \pi - {\sin}^{-} 1 \frac{- 1 - \sqrt{33}}{8} + 2 \pi k \\ {\sin}^{-} 1 \frac{- 1 + \sqrt{33}}{8} + 2 \pi k \\ \pi - {\sin}^{-} 1 \frac{- 1 + \sqrt{33}}{8} + 2 \pi k\end{matrix}\right.$ $\text{ }$ for integer $k$