The general value of theta satisfying sin^2 t | vedclass

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(C) Given the equation: $\sin^2 	heta + \sin 	heta = 2$Rearranging the terms,we get: $\sin^2 	heta + \sin 	heta - 2 = 0$Let $x = \sin 	heta$. The

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$n\pi + (-1)^n \frac{\pi}{2}$

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(C) Given the equation: $\sin^2 	heta + \sin 	heta = 2$Rearranging the terms,we get: $\sin^2 	heta + \sin 	heta - 2 = 0$Let $x = \sin 	heta$. Then the equation becomes $x^2 + x - 2 = 0$.Factoring the quadratic equation: $(x - 1)(x + 2) = 0$.This gives $x = 1$ or $x = -2$.Since the range of $\sin 	heta$ is $[-1, 1]$,$\sin 	heta$ cannot be $-2$.Therefore,$\sin 	heta = 1$.We know that $\sin 	heta = 1 = \sin(\frac{\pi}{2})$.The general solution for $\sin 	heta = \sin \alpha$ is $	heta = n\pi + (-1)^n \alpha$.Substituting $\alpha = \frac{\pi}{2}$,we get $	heta = n\pi + (-1)^n \frac{\pi}{2}$.

The general value of $\theta$ satisfying $\sin^2 \theta + \sin \theta = 2$ is

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