If $\cos 2 \theta=\sin \propto, \quad$ then $\theta=$
If $\cos 2 \theta=\sin \propto, \quad$ then $\theta=$
$2 n \pi \pm\left(\frac{\pi}{2}-\alpha\right), n \in z$
$n \pi \pm\left(\frac{\pi}{4}+\frac{\alpha}{2}\right), n \in z$
$\frac{1}{2}\left[n \pi+(-1)^{n} \propto\right], n \in z$
$n \pi \pm\left(\frac{\pi}{4}-\frac{\alpha}{2}\right), n \in z$
Solution
We have $\cos 2 \theta=\sin \alpha \Rightarrow \cos 2 \theta=\cos (90-\alpha)$ When $\cos \theta=\cos \alpha$, we get $\theta=2 n \pi \pm \alpha, n \in Z$ $\therefore \quad 2 \theta=2 \mathrm{n} \pi \pm(90-\alpha)$
$\theta=\frac{2 n \pi}{2} \pm\left(\frac{90-\alpha}{2}\right) \Rightarrow \theta=n \pi \pm\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)$