If $2^{\cos(x)}=\sin(x)$, where $0 <x < {\pi/2} $, Find $\sin(x)$.

Question

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

If $2^{\cos(x)}=\sin(x)$, where $0 <x < {\pi/2} $, Find $\sin(x)$.

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 1

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.

Answer 2

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.