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?
 A: By periodicity, we can restrict ourselves to $[0, 2\pi[$. As you've noticed, $x$ must be in $X:=[0, \pi/2]$.
One method to continue is to notice that, for $a$ in $[0,1]$, we have $\sqrt{a} \geq a$, therefore for $x$ in $X$ we have $$\sqrt{\sin(x)} + \sqrt{\cos(x)} \geq \sin(x) + \cos(x)$$
It is then easy to check that $\sin(x) + \cos(x) \geq 1$ for $x$ in $X$, for example by squaring. So, the domain of $f$ is $X$.
A: Hint: you must solve $$\sin(x)\geq 0$$ and $$\cos(x)\geq 0$$ and $$\sqrt{\sin(x)}+\sqrt{\cos(x)}-1\geq 0$$
A: 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
$x = \pi/2$.
A: You need $\sqrt{\sin(x)}+\sqrt{\sin(\dfrac{\pi}{2}-x)}\ge 1$. Looking at the graphics you can see that the domain is 
$$\left([0,\dfrac{\pi}{2}]\cup[2\pi,2\pi+\dfrac{\pi}{2}\cup\cdots\right)\cup\left([-2\pi,-2\pi+\dfrac{\pi}{2}]\cup[-4\pi,4\pi+\dfrac{\pi}{2}]\cup\cdots\right)$$ Finally the domain is the union for all $k\in\mathbb Z$ of the intervals
$$\left[2k\pi,2k\pi+\dfrac{\pi}{2}\right]$$
