The smallest positive angle which satisfies t | vedclass

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(A) Given equation: $2\sin^2 	heta + \sqrt{3} \cos 	heta + 1 = 0$Using $\sin^2 	heta = 1 - \cos^2 	heta$,we get:$2(1 - \cos^2 	heta) + \sqrt{3} \

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$\frac{5\pi}{6}$

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(A) Given equation: $2\sin^2 	heta + \sqrt{3} \cos 	heta + 1 = 0$Using $\sin^2 	heta = 1 - \cos^2 	heta$,we get:$2(1 - \cos^2 	heta) + \sqrt{3} \cos 	heta + 1 = 0$$2 - 2\cos^2 	heta + \sqrt{3} \cos 	heta + 1 = 0$$-2\cos^2 	heta + \sqrt{3} \cos 	heta + 3 = 0$$2\cos^2 	heta - \sqrt{3} \cos 	heta - 3 = 0$Let $x = \cos 	heta$. The quadratic equation is $2x^2 - \sqrt{3}x - 3 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{\sqrt{3} \pm \sqrt{3 - 4(2)(-3)}}{2(2)} = \frac{\sqrt{3} \pm \sqrt{3 + 24}}{4} = \frac{\sqrt{3} \pm \sqrt{27}}{4} = \frac{\sqrt{3} \pm 3\sqrt{3}}{4}$$x_1 = \frac{4\sqrt{3}}{4} = \sqrt{3}$ (Not possible as $|\cos 	heta| \le 1$)$x_2 = \frac{-2\sqrt{3}}{4} = -\frac{\sqrt{3}}{2}$So,$\cos 	heta = -\frac{\sqrt{3}}{2}$.The smallest positive angle $	heta$ is $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$.

The smallest positive angle which satisfies the equation $2\sin^2 \theta + \sqrt{3} \cos \theta + 1 = 0$ is

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