Domain of $f(x)=\sqrt{\sqrt{\sin x}+\sqrt{\cos x}-1}$ Find Domain of $f(x)=\sqrt{\sqrt{\sin x}+\sqrt{\cos x}-1}$
My try:
First of all $x$ belongs to First quadrant
Also  $$\sqrt{\sin x}+\sqrt{\cos x}-1 \ge 0$$  Squaring both sides we get
$$\sin x+\cos x+2\sqrt{\sin x\cos x} \ge 1$$
Any clue here?

For
$f(x)=\sqrt{\sqrt{\sin x}+\sqrt{\cos x}-1}$,
you first of all have to
restrict $x$ to the reals.
Then,
since $\sin$ and $\cos$
are periodic
with period $2\pi$,
you can restrict $x \in [0, 2 \pi)$.
Any restrictions there
are replicated in
$[2k\pi, 2(k+1)\pi)$.
Since you need
$\sin(x) \ge 0$,
this gives
$x \in [0, \pi]$.
Since you need
$\cos(x) \ge 0$,
this gives
$x \in [0, \pi/2]\cup [3\pi/2, 2\pi)$.
Combining these gives
$x \in [0, \pi/2]$.
Finally,
you want 
$\sqrt{\sin x}+\sqrt{\cos x}-1
\ge 0$.
If $f(x)
= \sqrt{\sin x}+\sqrt{\cos x}-1
$,
then
$f(x) = f(\pi/2-x)$,
so we only need to look at
$x \in [0, \pi/4]$.
$f(0) = 0$,
$f(\pi/4)
=\sqrt{2}-1
\gt 0$.
$\begin{array}\\
f'(x)
&=\frac12 \cos(x) \sin^{-1/2}(x)-\frac12 \sin(x) \cos^{-1/2}(x)\\
&=\frac12 \dfrac{\cos^{3/2}(x)- \sin^{3/2}(x)}{\sin^{1/2}(x) \cos^{1/2}(x)}\\
\end{array}
$
and since
$\cos(x) > \sin(x)$
for
$x \in [0, \pi/4)$,
$f'(x) > 0$ for
$0 \le x \lt \pi/4$
so $f(x)
\ge 0$
for
$0 \le x \le \pi/2$
with equality only at
$x=0$
and
$x = \pi/2$.
