The general solution of 2 sqrt{3} cos^2 theta | vedclass

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

Vedclass · Accepted Answer

$n \pi + (-1)^n \frac{\pi}{3}, n \in Z$

Vedclass · Answer

(A) Given equation: $2 \sqrt{3} \cos^2 	heta = \sin 	heta$ Using the identity $\cos^2 	heta = 1 - \sin^2 	heta$,we get: $2 \sqrt{3} (1 - \sin^2 	heta) = \sin 	heta$ $2 \sqrt{3} - 2 \sqrt{3} \sin^2 	heta = \sin 	heta$ $2 \sqrt{3} \sin^2 	heta + \sin 	heta - 2 \sqrt{3} = 0$ Let $x = \sin 	heta$. Then $2 \sqrt{3} x^2 + x - 2 \sqrt{3} = 0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{-1 \pm \sqrt{1^2 - 4(2 \sqrt{3})(-2 \sqrt{3})}}{2(2 \sqrt{3})} = \frac{-1 \pm \sqrt{1 + 48}}{4 \sqrt{3}} = \frac{-1 \pm 7}{4 \sqrt{3}}$ Case $1$: $x = \frac{6}{4 \sqrt{3}} = \frac{3}{2 \sqrt{3}} = \frac{\sqrt{3}}{2}$ Case $2$: $x = \frac{-8}{4 \sqrt{3}} = -\frac{2}{\sqrt{3}} \approx -1.15$ (Not possible as $|\sin 	heta| \leq 1$) So,$\sin 	heta = \frac{\sqrt{3}}{2} = \sin \frac{\pi}{3}$ The general solution is $	heta = n \pi + (-1)^n \frac{\pi}{3}, n \in Z$.

The general solution of $2 \sqrt{3} \cos^2 \theta = \sin \theta$ is

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