The number of solutions of cos 2 theta = sin | vedclass

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(B) Given the equation $\cos 2 	heta = \sin 	heta$. Using the identity $\cos 2 	heta = 1 - 2 \sin^2 	heta$,we have: $1 - 2 \sin^2 	heta = \sin

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$3$

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(B) Given the equation $\cos 2 	heta = \sin 	heta$. Using the identity $\cos 2 	heta = 1 - 2 \sin^2 	heta$,we have: $1 - 2 \sin^2 	heta = \sin 	heta$ $2 \sin^2 	heta + \sin 	heta - 1 = 0$ Factoring the quadratic equation: $2 \sin^2 	heta + 2 \sin 	heta - \sin 	heta - 1 = 0$ $2 \sin 	heta (\sin 	heta + 1) - 1 (\sin 	heta + 1) = 0$ $(\sin 	heta + 1)(2 \sin 	heta - 1) = 0$ This gives two cases: $1) \sin 	heta = -1 \Rightarrow 	heta = \frac{3 \pi}{2}$ $2) \sin 	heta = \frac{1}{2} \Rightarrow 	heta = \frac{\pi}{6}, \frac{5 \pi}{6}$ Since $	heta \in (0, 2 \pi)$,all three values $\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{3 \pi}{2}$ are valid solutions. Thus,the total number of solutions is $3$.

The number of solutions of $\cos 2 \theta = \sin \theta$ in $(0, 2 \pi)$ is

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