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:


Hint:
The solutions of $$\sqrt{7-4\sqrt2 \sin x}=2\cos(x)-\sqrt2 \tan(x)$$
are also solutions of 
\begin{align*}
7-4\sqrt{2}\sin x&=4\cos^2 x-4\sqrt{2}\cos x\tan x+2\tan^2 x\\
7-4\sqrt{2}\sin x&=4\cos^2 x-4\sqrt{2}\sin x+2\tan^2 x
\end{align*}
Last equation is equivalent to
$$4\cos^2 x+2\tan^2 x -7=0...(1)$$
Let $t=\cos^2 x$, so $(1)$ can be seen as
$$4t+2\left(\frac{1}{t}-1\right)-7=0\iff 4t^2-9t+2=0$$
which can be solved by the quadratic formula giving us $t\in\{2,\,\frac{1}{4}\}$. Since no real $x$ satisfies $\cos^2 x =2$ we take $\cos^2 x =\frac{1}{4}$.
