The solution set of the equation sin^2 theta | vedclass

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(B) Given equation: $\sin^2 	heta - \cos 	heta = \frac{1}{4}$ Using $\sin^2 	heta = 1 - \cos^2 	heta$,we get: $(1 - \cos^2 	heta) - \cos 	heta =

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$\left\{\frac{\pi}{3}, \frac{5\pi}{3}\right\}$

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(B) Given equation: $\sin^2 	heta - \cos 	heta = \frac{1}{4}$ Using $\sin^2 	heta = 1 - \cos^2 	heta$,we get: $(1 - \cos^2 	heta) - \cos 	heta = \frac{1}{4}$ $4 - 4\cos^2 	heta - 4\cos 	heta = 1$ $4\cos^2 	heta + 4\cos 	heta - 3 = 0$ Let $x = \cos 	heta$,then $4x^2 + 4x - 3 = 0$ $4x^2 + 6x - 2x - 3 = 0$ $2x(2x + 3) - 1(2x + 3) = 0$ $(2x - 1)(2x + 3) = 0$ So,$\cos 	heta = \frac{1}{2}$ or $\cos 	heta = -\frac{3}{2}$ (rejected as $-1 \le \cos 	heta \le 1$) For $\cos 	heta = \frac{1}{2}$ in $[0, 2\pi]$,$	heta = \frac{\pi}{3}$ or $	heta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$ Thus,the solution set is $\left\{\frac{\pi}{3}, \frac{5\pi}{3}ight\}$.

The solution set of the equation $\sin^2 \theta - \cos \theta = \frac{1}{4}$ in the interval $[0, 2\pi]$ is

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