# 2cos^2x-1=0,x=5pi/4?

May 3, 2018

#### Explanation:

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

The solution:

hit the two side by $\frac{1}{2}$

${\cos}^{2} \left(x\right) - \frac{1}{2} = 0$

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

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

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

$x = 45$ when $\cos x = \frac{1}{\sqrt{2}}$

$x = 315$ when $\cos x = \frac{1}{\sqrt{2}}$

$x = 135$ when $\cos x = - \frac{1}{\sqrt{2}}$

$x = 225$ when $\cos x = - \frac{1}{\sqrt{2}}$

or you mean

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

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

$x = \left(5 \frac{\pi}{4}\right) = 225$

${\cos}^{2} \left(225 - 1\right) =$

${\cos}^{2} \left(224\right) = 0.34127$