Question: I've been unable to solve this question since a long time 

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> If $2^{\cos(x)}=\sin(x)$, where $0 <x < {\pi/2} $, Find $\sin(x)$.
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I tried plotting the graph and found that there were two solutions, (one of them outside the domain, which is $\cos(x)=0$). I could not mathematically compute the second one. Would someone please help me to solve this problem?

Answer: 
For the range $0 \leq x \leq \frac \pi 2$, you already received the answer.

If it was $0 \leq x \leq \pi $, as you noticed, there is another root close to $x=2.5$ which can easily be found using Newton method the iterates of which being
$$\left(
\begin{array}{cc}
 n & x\_n \\
 0 & 2.500000000 \\
 1 & 2.543649683 \\
 2 & 2.542076822 \\
 3 & 2.542074833
\end{array}
\right)$$ which is not recognized by inverse symbolic calculators.
Square both sides of the equation and let $\cos(x)=a$,where $a\in[0,1]$ this gives us the equation $2^{2a}=1-a^2$. The equations agree at $a=0$, but the left hand side always has a positive derivative(is increasing), whereas the right hand side always has a negative derivative (is decreasing). Thus the function has no solution on a given interval (since $a=0$ corresponds to $x=\frac{\pi}{2}$). If you increase the interval to $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$ there will be another solution, and with a similar line of reasoning (by using the fact that the derivative of LHS will be greater than that of the RHS) you can show that there will only be 2 solutions, but I'm not sure how you would find the second one.