QJEEM22Maed73: JEE Main 2022 Mathematics - Class 11 JEEM22Maed73 Solution: The number of values of x in the interval for which holds, is…

${{14} \over {{{\sin }^2}x}} - 2{\sin ^2}x = 21 - 4(1 - {\sin ^2}x)$

Let ${\sin ^2}x = t$

$\Rightarrow 14 - 2{t^2} = 21t - 4t + 4{t^2}$

$\Rightarrow 6{t^2} + 17t - 14 = 0$

$\Rightarrow 6{t^2} + 21t - 4t - 14 = 0$

$\Rightarrow 3t(2t + 7) - 2(2t + 7) = 0$

$\Rightarrow {\sin ^2}x = {2 \over 3}$ or $- {7 \over 3}$ (rejected)

$\Rightarrow \sin x = \, \pm \,\sqrt {{2 \over 3}}$

$\therefore$ $\sin x = \, \pm \,\sqrt {{2 \over 3}}$ has 4 solutions in $\left( {{\pi \over 4},\,{{7\pi } \over 4}} \right)$