Trying to solve $\sqrt{2\cos^2(x)-\sqrt{3}}+\sqrt2 \sin(x)=0$ The equation is
$$\sqrt{2\cos^2(x)-\sqrt{3}}+\sqrt2 \sin(x)=0$$
I solve it thus:
$$
\begin{cases}
2\cos^2(x)-\sqrt3=2\sin^2(x) \\
-\sqrt2 \sin(x)\ge 0 \iff \sin(x)\le 0
\end{cases}
$$
The first equation boils down to
$$2\cos^2(x)=2(1-\cos^2(x))+\sqrt3$$
$$4cos^2(x)-2=\sqrt3$$
$$2(2\cos^2(x)-1)=\sqrt3$$
$$2\cos(2x)=\sqrt3$$
$$\cos(2x)=\frac{\sqrt3}{2}$$
$$2x=\pm \arccos(\frac{\sqrt3}{2})+2n\pi$$
$$2x=\pm \frac{\pi}{6}+2n\pi$$
$$x=\pm \frac{\pi}{12}+n\pi$$
Considering the condition $\sin(x)\le 0$, we are left with
$$x=- \frac{\pi}{12}+n\pi$$
But the textbook's answer is
$$x=- \frac{\pi}{12}+2n\pi; x=- \frac{11\pi}{12}+2n\pi$$
What did I overlook?

P.S. The problem and the answer from the texbook:

 A: $$2\sin^2x=2\cos^2x-\sqrt3$$
$$\iff2\sin^2x=2(1-\sin^2x)-\sqrt3$$
$$\iff\sin^2x=\dfrac{2-\sqrt3}4=\dfrac{(\sqrt3-1)^2}{8}$$
As $\sin x<0,\sin x=-\dfrac{\sqrt3-1}{2\sqrt2}=\dfrac12\cdot\dfrac1{\sqrt2}-\dfrac{\sqrt3}2\cdot\dfrac1{\sqrt2}=\sin\left(\dfrac\pi6-\dfrac\pi4\right)$
$x=n\pi+(-1)^n\left(\dfrac\pi6-\dfrac\pi4\right)$ where $n$ is any integer
A: As Brent points out in his comment, your only mistake was applying the condition $\sin x<0$ at the end:
1) If $x=\frac{\pi}{12}+n\pi$, $\;\;\sin x>0$ for $n$ even and $\sin x<0$ for $n$ odd, so this gives
$\hspace{.4 in} x=\frac{\pi}{12}+(2n+1)\pi=\frac{13\pi}{12}+2n\pi=-\frac{11\pi}{12}+2n\pi$
2) If $x=-\frac{\pi}{12}+n\pi$, $\;\;\sin x<0$ for $n$ even and $\sin x>0$ for $n$ odd, so this gives
$\hspace{.4 in} x=-\frac{\pi}{12}+2n\pi=\frac{23\pi}{12}+2n\pi$
A: Using $\cos2y=1-2\sin^2y=2\cos^2y-1$ on
$$2\sin^2x=2\cos^2x-\sqrt3$$
$$\iff1-\cos2x=1+\cos2x-\sqrt3\iff\cos2x=\dfrac{\sqrt3}2=\cos30^\circ$$
$$2x=360^\circ n\pm30^\circ\iff x=180^\circ n\pm15^\circ$$ where $n$ in any integer
Case$\#1:$ $+\implies x=180^\circ n+15^\circ$
But as $\sin x<0,x$ lies in the third or in the fourth quadrant
$\implies n$ must be odd $=2m+1$(say)
Case$\#2:$ $-\implies x=180^\circ n-15^\circ$
For the reason mentioned in Case$\#1,$ here $n$ must be even $=2m$(say)
A: Take advantage of the fact that $\cos^2{x}=1-\sin^2{x}$. Then solve for $\sin{x}$, then solve for $x$. $\frac{\pi}{12}=15^o$
