<s>[INST] Solving $\cos(2\theta)=\sin\left(\frac{\theta}{2}\right)$ Consider an acute angle $\theta$, which has the property that
$$\cos(2\theta)=\sin\left(\frac{\theta}{2}\right).$$
I am trying to find the value of $\theta$. Using the identity
$$\cos(2\theta)=\sin\left(\frac{\pi}{2}-2\theta\right),$$
I can write the first equation as
$$\sin\left(\frac{\pi}{2}-2\theta\right)=\sin\left(\frac{\theta}{2}\right).$$
Equating the argument of the $\sin$ functions, we see that
\begin{align}
\frac{\pi}{2}-2\theta&=\frac{\theta}{2} \\
\frac{5}{2}\theta&=\frac{\pi}{2} \\
\theta&=\frac{\pi}{5}.
\end{align}
While this yields the correct answer, I do not know how we can justify equating the arguments of the $\sin$ functions.
 [/INST] Note that the solution to $$\sin(x) = \sin(y)$$ is $$x = 2k\pi + y \text{ and } x = 2k\pi + \pi - y$$ So in this case we have $$\frac\pi 2 - 2\theta = 2k\pi  + \frac\theta 2 \text{ and } \frac\pi 2 - 2\theta = 2k\pi  + \pi - \frac\theta 2$$The solutions are $$\frac{5\theta}{2} = \frac \pi 2 - 2k\pi \implies \theta = \frac \pi 5 - \frac{4k\pi}{5}, \ \ k \in \mathbb{Z} \\ \frac{3\theta}{2} = -\frac \pi 2 - 2k\pi \implies \theta = -\frac \pi 3 - \frac{4k\pi}{3}, \ \ k \in \mathbb{Z}$$
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