0=2cos^2x + 2*sqrt2 * cos x + 1
When I tried the quadratic formula, the discriminant was equal to zero. So i know this can be factored, but i do not know how to deal with sqrt2.
Thank you for your time
$\displaystyle \displaystyle \begin{align*} 0 &= 2\cos^2{(x)} + 2\sqrt{2}\cos{(x)} + 1 \\ 0 &= \cos^2{(x)} + \sqrt{2}\cos{(x)} + \frac{1}{2} \\ 0 &= \cos^2{(x)} + \sqrt{2}\cos{(x)} + \left( \frac{\sqrt{2}}{2} \right) ^2 - \left( \frac{\sqrt{2}}{2} \right) ^2 + \frac{1}{2} \\ 0 &= \left[ \cos{(x)} + \frac{\sqrt{2}}{2} \right] ^2 - \frac{1}{2} + \frac{1}{2} \\ 0 &= \left[ \cos{(x)} + \frac{\sqrt{2}}{2} \right] ^2 \end{align*}$