The number of values of x in the interval lef | vedclass

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(D) Given equation: $14 \operatorname{cosec}^{2} x - 2 \sin^{2} x = 21 - 4 \cos^{2} x$ Substitute $\cos^{2} x = 1 - \sin^{2} x$: $14 \operatorname{cos

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$4$

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(D) Given equation: $14 \operatorname{cosec}^{2} x - 2 \sin^{2} x = 21 - 4 \cos^{2} x$ Substitute $\cos^{2} x = 1 - \sin^{2} x$: $14 \operatorname{cosec}^{2} x - 2 \sin^{2} x = 21 - 4(1 - \sin^{2} x)$ $14 \operatorname{cosec}^{2} x - 2 \sin^{2} x = 17 + 4 \sin^{2} x$ $14 \operatorname{cosec}^{2} x - 6 \sin^{2} x = 17$ Let $\sin^{2} x = p$,then $\operatorname{cosec}^{2} x = \frac{1}{p}$: $\frac{14}{p} - 6p = 17 \implies 14 - 6p^{2} = 17p$ $6p^{2} + 17p - 14 = 0$ $(6p - 4)(p + 3.5) = 0$ (or using quadratic formula): $p = \frac{2}{3}$ or $p = -3.5$ (rejected as $\sin^{2} x \geq 0$) So,$\sin^{2} x = \frac{2}{3} \implies \sin x = \pm \sqrt{\frac{2}{3}}$ In the interval $\left(\frac{\pi}{4}, \frac{7 \pi}{4}\right)$,$\sin x$ takes the value $\sqrt{\frac{2}{3}}$ twice and $-\sqrt{\frac{2}{3}}$ twice. Thus,there are $4$ solutions.

The number of values of $x$ in the interval $\left(\frac{\pi}{4}, \frac{7 \pi}{4}\right)$ for which $14 \operatorname{cosec}^{2} x - 2 \sin^{2} x = 21 - 4 \cos^{2} x$ holds,is

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