(C) Given $3 \operatorname{cosec} x = 4 \sin x$ $\Rightarrow \frac{3}{\sin x} = 4 \sin x$ $\Rightarrow 4 \sin^2 x = 3$ $\Rightarrow \sin^2 x = \frac{3

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(C) Given $3 \operatorname{cosec} x = 4 \sin x$ $\Rightarrow \frac{3}{\sin x} = 4 \sin x$ $\Rightarrow 4 \sin^2 x = 3$ $\Rightarrow \sin^2 x = \frac{3}{4}$ $\Rightarrow \sin x = \pm \frac{\sqrt{3}}{2}$ Since $\sin x = \frac{\sqrt{3}}{2}$ at $x = \frac{\pi}{3}$ and $\sin x = -\frac{\sqrt{3}}{2}$ at $x = -\frac{\pi}{3}$,the values of $x$ are $\pm \frac{\pi}{3}$.

Class 11 Mathematics Trigonometrical Equations Solution of trigonometrical equations

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The values of $x$ satisfying the equation $3 \operatorname{cosec} x = 4 \sin x$ are

AP EAMCET 2021 All AP EAMCET

A
$\frac{\pi}{6}, \frac{\pi}{3}$
B
$\pm \frac{\pi}{6}$
C
$\pm \frac{\pi}{3}$
D
$\frac{\pi}{3}, \frac{\pi}{4}$

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