Find the range of $f(x) = \sqrt{\log(\cos(\sin x))}$ 
Find the range of the function
  $$f(x)=\sqrt{\log(\cos(\sin(x))}.$$

What I did was:
$$\log(\cos(\sin(x))) \geq 0$$
Then 
$$\log_e\cos(\sin(x)) \geq 0$$
The next step of the problem is 
$$\cos(\sin(x)) \geq e^0$$
This is the part I didn’t understand. Please explain it to me.
 A: Note that $\sqrt{\log(\cos(\sin(x)))}$ is defined if and only if  $\cos(\sin(x))\geq 1$ otherwise we have the square root of a negative number. Now since the cosine function is upper bounded by $1$, $\cos(\sin(x))$ has to be $1$ (such equality holds for example when $x=0$). So what is the range of the given function?
A: Okay.  $e > 1$.  
So if $x > 0$ then $e^x > 1$ and $e^0 = 1$ and if $y < 0$ then $e^{y} = e^{-|y|} = \frac 1{e^{|y|} }< 1$ and so $0 < e^y < 1$.
So $e^x > 0$ always.  $0 < e^x < 1$ if and only if $x < 0$. and $e^x = 1$ if and only if $x = 0$ and $e^x > 1$ if and only if $x > 0$.
Claim:   $a < b$ if and only if $e^a < e^b$.
Pf:  If $a < b$ then $b-a > 0$ so $e^{b-a} > 1$ and so 
$e^b = e^{a + (b-a)} = e^a*e^{b-a} > e^a* 1 = e^a$.
If $e^a < e^b$ then $1 < \frac {e^b}{e^a} = e^{b-a}$.  So $b-a > 0$ and $b > a$.
Claim 2:  If $a > 0$ and $b > 0$, then $a < b \iff \ln a < \ln b$.
Pf:
If $a > 0$ then $a = e^{\ln a}$ and if $b > 0$ then $b = e^{\ln b}$.
So $a < b$ if and only if $e^{\ln a} < e^{\ln b}$ if and only if $\ln a < \ln b$
.......
And now we can do your problem.
$\sqrt{\ln(\cos(\sin x))} $ only exists if 
$\ln(\cos(\sin x)) \ge 0$.  This occurs if and only if
$\cos(\sin x) \ge e^0 = 1$.
But $-1 \le \cos w \le 1$ so the only way this can happen is if $\cos(\sin x) = 1$.
Now $\cos w = 1$ can only happen if $w = 0$.
So $\sin x = 0$.  That occurs if $x = 0$ or $x = 180^{\circ}$.
So ....
$x = 0$ or $180^{\circ}$
$\sin x = 0$
$\cos(\sin x) = 1$
$\log (\cos(\sin x)) = 0$
$\sqrt{\log (\cos(\sin x))} = 0$ 
and the range is $\{0\}$.
The end.
BTW was $\log$ suppose to be $\log_{10}$ or $\log_e = \ln$?  It doesn't actually matter.  THe answer will be the same.
