# How do you solve the identity  2 sin^2 x = 2 + cos x?

Aug 6, 2015

$x = \left(2 n - 1\right) \setminus \frac{\setminus \pi}{2} \text{ or " x = (2n-1)\pi \pm {\pi}/3 , " } n \in \mathbb{Z}$

#### Explanation:

${\sin}^{2} x + {\cos}^{2} x \equiv 1$
$\setminus \therefore 2 {\sin}^{2} x = 2 - 2 {\cos}^{2} x$

Substituting above to original equation:
$2 - 2 {\cos}^{2} x = 2 + \cos x$

Re-arrange:
$2 {\cos}^{2} x + \cos x = 0$
$\cos x \left(2 \cos x + 1\right) = 0$

$\cos x = 0 \text{ or } \cos x = - \setminus \frac{1}{2}$

$x = \left(2 n - 1\right) \setminus \frac{\setminus \pi}{2} \text{ or " x = (2n-1)\pi \pm {\pi}/3 , " } n \in \mathbb{Z}$