Trying to solve $\sqrt{7-4\sqrt2 \sin x}=2\cos(x)-\sqrt2 \tan(x)$ The equation is
$$\sqrt{7-4\sqrt2 \sin x}=2\cos(x)-\sqrt2 \tan(x)$$
We get the system
$$
\begin{cases}
7-4\sqrt 2 \sin(x)=4\cos^2(x)-2\sqrt2\cos(x)\tan(x)+2\tan^2(x) \\
2\cos(x)-\sqrt2 \tan(x)\ge 0
\end{cases}
$$
I transformed the equation thus:
$$7(\sin^2(x)+cos^2(x))-4\sqrt 2 \sin(x)=4\cos^2(x)-2\sqrt2\sin(x)+2\tan^2(x)$$
$$7\sin^2(x)+3cos^2(x)-2\sqrt 2 \sin(x)-2\frac{\sin^2(x)}{1-sin^2(x)}=0$$
I multiply the whole equation by $(1-sin^2(x))$ and then substitute sin(x) with t:
$$4t^4-2\sqrt2 t^3+t^2+2\sqrt2 t - 3 =0$$
And here I'm stuck. The polynomial is seemingly non-factorizable. 
A hint would be welcome. (0:

P.S. The problem as it is presented in the texbook:


$$2\cos x-\sqrt2\tan x=\dfrac{2-2\sin^2x-\sqrt2\sin x}{\cos x}=-\sqrt2\cdot\dfrac{\sqrt2\sin^2x+\sin x-\sqrt2}{\cos x}$$
Now $\sqrt2\sin^2x+\sin x-\sqrt2=(\sqrt2\sin x-1)(\sin x+\sqrt2)$ and $\sin x+\sqrt2\ge\sqrt2-1>0$
So we need $\dfrac{\sqrt2\sin x-1}{\cos x}\le0$
If $\cos x>0, \sqrt2\sin x-1\le0\iff\sin x\le\dfrac1{\sqrt2}$
$2n\pi\le x\le 2n\pi+\dfrac\pi4\  \ \ \  (1A)\ \ $  OR $\ \ 2n\pi+\dfrac{3\pi}2<x\le2n\pi+2\pi\  \ \ \  (1B)$
$(1A),(1B)$ can be clubbed as $2n\pi-\dfrac\pi2< x\le 2n\pi+\dfrac\pi4\  \ \ \  (2)$
Else if $\cos x<0, \sqrt2\sin x-1\ge0\iff\sin x\ge\dfrac1{\sqrt2}$
$2n\pi+\dfrac\pi2\le x\le 2n\pi+\dfrac{3\pi}4\  \ \ \  (3)$
Now from the other answer, $$\cos2x=2\cos^2x-1=-\dfrac12=\cos\left(\pi-\dfrac\pi3\right)$$
$$2x=2m\pi\pm\dfrac{2\pi}3\iff x=m\pi\pm\dfrac\pi3$$
Can you check which values of $x$ conform to $(1A),(1B)$ or $(2)$  and $(3)$
