JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Trigonometric Equations

Starting with the given equation $14{\csc }^{2}x-2{\sin }^{2}x=21-4{\cos }^{2}x$, we rewrite the right side using ${\cos }^{2}x=1-{\sin }^{2}x$:

$21-4(1-{\sin }^{2}x)=17+4{\sin }^{2}x$

So the equation becomes:
$14{\csc }^{2}x-2{\sin }^{2}x=17+4{\sin }^{2}x$

Rearranging: $14{\csc }^{2}x-6{\sin }^{2}x=17$

Substituting ${\csc }^{2}x=\frac{1}{{\sin }^{2}x}$ and letting $p={\sin }^{2}x$:
$\frac{14}{p}-6p=17$

Multiplying by $p$: $14-6p^2=17p$

Rearranging: $6p^2+17p-14=0$

Using the quadratic formula: $p=\frac{-17\pm \sqrt{289+336}}{12}=\frac{-17\pm 25}{12}$

This gives $p=\frac{2}{3}$ or $p=-\frac{7}{2}$ (rejected since $p={\sin }^{2}x\ge 0$).

So ${\sin }^{2}x=\frac{2}{3}$, which means $\sin x=\pm \sqrt{\frac{2}{3}}$.

In the interval $\left(\frac{\pi }{4},\frac{7\pi }{4}\right)$, each of the two values $\sin x=\sqrt{\frac{2}{3}}$ and $\sin x=-\sqrt{\frac{2}{3}}$ occurs twice (once in each relevant quadrant), giving a total of 4 solutions.