If 2sin^2 theta = 3cos theta,where 0 le theta | vedclass

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(B) Given the equation: $2\sin^2 	heta = 3\cos 	heta$Using the identity $\sin^2 	heta = 1 - \cos^2 	heta$,we get:$2(1 - \cos^2 	heta) = 3\cos 	h

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

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(B) Given the equation: $2\sin^2 	heta = 3\cos 	heta$Using the identity $\sin^2 	heta = 1 - \cos^2 	heta$,we get:$2(1 - \cos^2 	heta) = 3\cos 	heta$$2 - 2\cos^2 	heta = 3\cos 	heta$$2\cos^2 	heta + 3\cos 	heta - 2 = 0$Let $x = \cos 	heta$. Then $2x^2 + 3x - 2 = 0$.Solving the quadratic equation using the quadratic formula:$x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-2)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm 5}{4}$This gives two values for $x$:$x = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{1}{2}$ or $x = \frac{-3 - 5}{4} = -2$Since $-1 \le \cos 	heta \le 1$,we discard $x = -2$.Thus,$\cos 	heta = \frac{1}{2}$.In the interval $0 \le 	heta \le 2\pi$,$\cos 	heta = \frac{1}{2}$ at $	heta = \frac{\pi}{3}$ and $	heta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$.

If $2\sin^2 \theta = 3\cos \theta$,where $0 \le \theta \le 2\pi$,then $\theta = $

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