<s>[INST] Solve the following equation: $\sqrt {\sin x - \sqrt {\cos x + \sin x} } = \cos x$ Solve the following equation: 
\begin{array}{l}{\sqrt{\sin x-\sqrt{\cos x+\sin x}}=\cos x} \\ \text{my try as follows:}\\{\sin x-\sqrt{\cos x+\sin x}=\cos ^{2} x} \\ {\sin x-\cos ^{2} x=\sqrt{\cos x+\sin x}} \\ {\sin ^{2} x+\cos ^{4} x-2 \sin x \cos ^{2} x=\cos x+\sin x} \\ {\sin ^{2} x+\cos ^{4} x-2 \sin x \cos ^{2} x-\cos x-\sin x=0} \\ {\sin ^{2} x+\cos ^{4} x-2 \sin x\left(1-\sin ^{2} x\right)-\cos x-\sin x=0} \\{\sin ^2}x + {\left( {1 - {{\sin }^2}x} \right)^2} - 2\sin x\left( {1 - {{\sin }^2}x} \right) - \cos x - \sin x = 0\\ {\sin ^{2} x+\sin ^{4} x-2 \sin ^{2} x+1-2 \sin x+2 \sin ^{3} x-\cos x-\sin x=0} \\ {\sin ^{4} x+2 \sin ^{3} x-\sin ^{2} x-3\sin x-\cos x+1=0}\end{array}
Now i think it gets more complicated , any help would be appreciated 
 [/INST] We see that $\sin{x}\geq0$ and $\cos{x}\geq0.$
Let $f(x)=\sqrt{\sin{x}-\sqrt{\sin{x}+\cos{x}}},$ where $x\in\left[0,\frac{\pi}{2}\right].$
Thus, $$f'(x)=\frac{\cos{x}-\frac{\cos{x}-\sin{x}}{2\sqrt{\sin{x}+\cos{x}}}}{2\sqrt{\sin{x}-\sqrt{\sin{x}+\cos{x}}}}>0$$ for $\cos{x}-\sin{x}<0.$
But for $\cos{x}-\sin{x}\geq0$ we obtain:
$$f'(x)=\frac{4\cos^2x(\sin{x}+\cos{x})-(\cos{x}-\sin{x})^2}{4\sqrt{\sin{x}+\cos{x}}\sqrt{\sin{x}-\sqrt{\sin{x}+\cos{x}}}(2\cos{x}\sqrt{\sin{x}+\cos{x}}+\cos{x}-\sin{x})}$$ and
$$4\cos^2x(\sin{x}+\cos{x})-(\cos{x}-\sin{x})^2=4\cos^2x(\sin{x}+\cos{x})+2\sin{x}\cos{x}-1=$$
$$=4\cos^2x(\sin{x}+\cos{x})+2\sin{x}\cos{x}-\sqrt{\sin^2x+\cos^2x}\geq$$
$$\geq2\cos^2x(\sin{x}+\cos{x})-\sin{x}-\cos{x}=(\sin{x}+\cos{x})^2(\cos{x}-\sin{x})\geq0,$$
which says that $f$ increases.
But $\cos$ decreases, which says that there is unique $\alpha\in\left[0,\frac{\pi}{2}\right]$, for which our equation has root.
$\frac{\pi}{2}$ is a root, which gives the answer:
$$\left\{\frac{\pi}{2}+2\pi k|k\in\mathbb Z\right\}$$
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