The general solution of the equation sin^2 th | vedclass

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(B) Given equation: $\sin^2 	heta + 3 \cos^2 	heta = 5 \sin 	heta$ Substitute $\cos^2 	heta = 1 - \sin^2 	heta$: $\sin^2 	heta + 3(1 - \sin^2

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$n \pi + (-1)^n \frac{\pi}{6}, n \in Z$

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(B) Given equation: $\sin^2 	heta + 3 \cos^2 	heta = 5 \sin 	heta$ Substitute $\cos^2 	heta = 1 - \sin^2 	heta$: $\sin^2 	heta + 3(1 - \sin^2 	heta) = 5 \sin 	heta$ $\sin^2 	heta + 3 - 3 \sin^2 	heta = 5 \sin 	heta$ $-2 \sin^2 	heta - 5 \sin 	heta + 3 = 0$ $2 \sin^2 	heta + 5 \sin 	heta - 3 = 0$ Factor the quadratic equation: $(2 \sin 	heta - 1)(\sin 	heta + 3) = 0$ This gives $\sin 	heta = \frac{1}{2}$ or $\sin 	heta = -3$. Since $-1 \le \sin 	heta \le 1$,we discard $\sin 	heta = -3$. For $\sin 	heta = \frac{1}{2} = \sin \frac{\pi}{6}$,the general solution is $	heta = n \pi + (-1)^n \frac{\pi}{6}, n \in Z$.

The general solution of the equation $\sin^2 \theta + 3 \cos^2 \theta = 5 \sin \theta$ is

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