If cot theta = sin 2theta (where theta neq np | vedclass

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(B) Given the equation: $\cot 	heta = \sin 2	heta$Since $\cot 	heta = \frac{\cos 	heta}{\sin 	heta}$ and $\sin 2	heta = 2 \sin 	heta \cos 	het

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$45^o$ and $90^o$

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(B) Given the equation: $\cot 	heta = \sin 2	heta$Since $\cot 	heta = \frac{\cos 	heta}{\sin 	heta}$ and $\sin 2	heta = 2 \sin 	heta \cos 	heta$,we have:$\frac{\cos 	heta}{\sin 	heta} = 2 \sin 	heta \cos 	heta$$\cos 	heta = 2 \sin^2 	heta \cos 	heta$$\cos 	heta (1 - 2 \sin^2 	heta) = 0$This implies either $\cos 	heta = 0$ or $1 - 2 \sin^2 	heta = 0$.Case $1$: $\cos 	heta = 0 \implies 	heta = 90^o$ (within the principal range).Case $2$: $2 \sin^2 	heta = 1 \implies \sin^2 	heta = \frac{1}{2} \implies \sin 	heta = \pm \frac{1}{\sqrt{2}}$.For $\sin 	heta = \frac{1}{\sqrt{2}}$,$	heta = 45^o$.Thus,the values are $45^o$ and $90^o$.

If $\cot \theta = \sin 2\theta$ (where $\theta \neq n\pi$,$n$ is an integer),then $\theta$ is equal to:

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