If cos 2theta = sin alpha,then theta = | vedclass

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(D) Given $\cos 2	heta = \sin \alpha$. We know that $\sin \alpha = \cos(\frac{\pi}{2} - \alpha)$. So,$\cos 2	heta = \cos(\frac{\pi}{2} - \alpha)$. T

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$n\pi \pm (\frac{\pi}{4} - \frac{\alpha}{2}), n \in \mathbb{Z}$

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(D) Given $\cos 2	heta = \sin \alpha$. We know that $\sin \alpha = \cos(\frac{\pi}{2} - \alpha)$. So,$\cos 2	heta = \cos(\frac{\pi}{2} - \alpha)$. The general solution for $\cos x = \cos y$ is $x = 2n\pi \pm y$,where $n \in \mathbb{Z}$. Applying this,$2	heta = 2n\pi \pm (\frac{\pi}{2} - \alpha)$. Dividing by $2$,we get $	heta = n\pi \pm (\frac{\pi}{4} - \frac{\alpha}{2})$,where $n \in \mathbb{Z}$.

If $\cos 2\theta = \sin \alpha$,then $\theta =$

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