The general solution of the equation 3 sec^2 | vedclass

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Question

(C) Given equation: $3 \sec^2 	heta = 2 \operatorname{cosec} 	heta$ $\Rightarrow \frac{3}{\cos^2 	heta} = \frac{2}{\sin 	heta}$ $\Rightarrow \frac

Vedclass · Accepted Answer

$n \pi + (-1)^n \frac{\pi}{6}, n \in Z$

Vedclass · Answer

(C) Given equation: $3 \sec^2 	heta = 2 \operatorname{cosec} 	heta$ $\Rightarrow \frac{3}{\cos^2 	heta} = \frac{2}{\sin 	heta}$ $\Rightarrow \frac{3}{1 - \sin^2 	heta} = \frac{2}{\sin 	heta}$ $\Rightarrow 3 \sin 	heta = 2 - 2 \sin^2 	heta$ $\Rightarrow 2 \sin^2 	heta + 3 \sin 	heta - 2 = 0$ $\Rightarrow (2 \sin 	heta - 1)(\sin 	heta + 2) = 0$ Since $\sin 	heta$ cannot be $-2$,we have $\sin 	heta = \frac{1}{2} = \sin \frac{\pi}{6}$. The general solution for $\sin 	heta = \sin \alpha$ is $	heta = n \pi + (-1)^n \alpha$. Therefore,$	heta = n \pi + (-1)^n \frac{\pi}{6}, n \in Z$.

The general solution of the equation $3 \sec^2 \theta = 2 \operatorname{cosec} \theta$ is

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