# The value pi/16 is a solution for the equation 2 cos^2(4x)-1=0. True or False?

#### realbrandon

The value pi/16 is a solution for the equation 2 cos^2(4x)-1=0. True or False?

I think it is true because: 2cos^2(π/4)-1= 2(√2/2)^2-1= 2(1/2)-1= 1-1= 0

Is it TRUE please explain. Thank you!

#### skeeter

MHF Helper
Your evaluation works ... why the doubt?

1 person

#### Prove It

MHF Helper
Yes it's true for the exact reason you posted. You could also try to solve the equation...

\displaystyle \begin{align*} 2\cos^2{(4x)} - 1 &= 0 \\ 2\cos^2{(4x)} &= 1 \\ \cos^2{(4x)} &= \frac{1}{2} \\ \cos{(4x)} &= \pm \frac{1}{\sqrt{2}} \\ 4x &= \left\{ \frac{\pi}{4} , \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \right\} + 2\pi\,n \textrm{ where } n \in \mathbf{Z} \\ x &= \left\{ \frac{\pi}{16}, \frac{3\pi}{16}, \frac{5\pi}{16}, \frac{7\pi}{16} \right\} + \frac{\pi}{2}n \end{align*}

Clearly \displaystyle \begin{align*} x = \frac{\pi}{16} \end{align*} is a solution.

1 person