# If 2cos^2x - 2sin^2x = 1 , what is the value of x?

Mar 7, 2016

$x = \pm \frac{\pi}{6} + n \pi$ for $n \in \mathbb{Z}$

#### Explanation:

Using the identity $\cos \left(2 x\right) = {\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right)$ we have

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

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

$= 2 \cos \left(2 x\right)$

$\implies \cos \left(2 x\right) = \frac{1}{2}$

$\implies 2 x = \pm \frac{\pi}{3} + 2 n \pi$ for $n \in \mathbb{Z}$

$\therefore x = \pm \frac{\pi}{6} + n \pi$ for $n \in \mathbb{Z}$