How can I solve this trig exponential equation ??

Question

$$(\sin x)^{(\log_2 \sin x)^2+\log_2(1-\cos2x)} = 1$$

I tried taking $\log_2$ of both sides and I got:

$(\sin(x))^2\cdot(1-\cos(x))=1$

Is it correct? If it is, how should I continue?

Answer 1

$[(\log_2(\sin(x)))^2+\log_2(1-\cos(2x))] \ln(\sin(x)) =0$

So, either $\ln(\sin(x))=0$ and $\sin(x)=1$, or

\begin{align} (\log_2(\sin(x)))^2+\log_2(1-1+2\sin^2(x)) &=0 \\ (\log_2(\sin(x)))^2+\log_2(2\sin^2(x)) &= 0 \\ (\log_2(\sin(x)))^2+2\log_2(\sin(x))+1 &= 0 \end{align}

Let $m=\log_2(\sin(x))$, $\ldots$

\begin{align} m^2+2m+1 &= 0 \\ (m+1)^2 &= 0 \\ m &= -1 \\ \sin(x) &= \frac{1}{2} \end{align}

So, the solutions are when $\sin(x)=1$ or when $\sin(x)=\frac{1}{2}$. Use reference angles, and be cognizant about which angles work in the original equation. (For example, you can’t take a logarithm of a negative number over the reals.)

Answer 2

Suppose $(\sin x)^{(\log_2 \sin x)^2+\log_2(1-\cos(2x))} = 1 $.

Then either $\sin(x) = 1$ or $(\log_2 \sin x)^2+\log_2(1-\cos(2x)) = 0 $.

If $\sin(x) = 1$, then $x = \pi(2k+1)$ where $k$ is an integer.

If $(\log_2 \sin x)^2+\log_2(1-\cos(2x)) = 0 $, then

$\begin{array}\\ 0 &=(\log_2 \sin x)^2+\log_2(1-\cos(2x))\\ &=(\log_2 \sin x)^2+\log_2(2\sin^2(x))\\ &=(\log_2 \sin x)^2+\log_2(2)+2\log_2(\sin(x))\\ &=v^2+\log_2(2)+2v \qquad\text{where } v = \log_2(\sin(x))\\ &=v^2+2v+1-1+\log_2(2)\\ &=(v+1)^2+\log_2(2)-1\\ &=(v+1)^2 \qquad\text{since }\log_2(2) = 1\\ \end{array} $

Therefore $v+1 = 0$ or $\log_2(\sin(x)) = -1 $ or $\sin(x) =\frac12 $ or $x =2k\pi+\pi/6 $ and $x =2k\pi+5\pi/6 $ (as pointed out by egreg) where $k$ is an integer.