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.

$$
\begin{aligned}
\operatorname{cos} 2 \theta &=\sin \frac{\theta}{2}=\cos \left(\frac{\pi}{2}-\frac{\theta}{2}\right) \\
2 \theta &=2 n \pi \pm\left(\frac{\pi}{2}-\frac{\theta}{2}\right) \\
\frac{5 \theta}{2} &=2 n \pi+\frac{\pi}{2} \text { or } \frac{3 \theta}{2}=2 n \pi-\frac{\pi}{2} \\
\theta &=\frac{(4 n+1) \pi}{5} \text { or } \frac{(4 n-1) \pi}{3},
\end{aligned}
$$
where $n\in Z.$
