Is my solution correct? (trigonometry) $$2\cos x\left(\cos x-\sqrt{8}\tan x\right)<5$$
After expanding I get $2\sin^2 x+4\sqrt{2}\sin x-5>0$. I then factorized $(2\sin x+3\sqrt{2})(2\sin x+\sqrt{2})>0$. The first one is always positive so $\sin x>\sin(-\pi/4)$. How to continue?

$\sin x>-\sqrt{2}/2$ and you have $\sin(-\pi/4)=\sin(5\pi/4)=-\sqrt{2}/2$. The big arc between $-\pi/4$ and $5\pi/4$ is the solution :
$$\left[\frac{-\pi}{4}, \frac{5\pi}{4}\right]~~~\text{(mod }2\pi\text{)}$$
Another way to denote this solution is 
$$\bigcup\limits_{n\in\mathbb{Z}}\left[\frac{-\pi}{4}+2n\pi, \frac{5\pi}{4}+2n\pi\right]$$
