JEE Main 2022 · 25 Jun · Shift 1, Mathematics question 25

In $s=\sin^2x$. $\dfrac{14}s-2s=21-4(1-s)$ gives $6s^2+17s-14=0$, so $s=\dfrac23$.

Solutions. $\sin x=\pm\sqrt{2/3}$, with $\sin^{-1}\sqrt{2/3}\approx0.955>\frac\pi4$. The four solutions in $[0,2\pi)$, about 0.955, 2.186, 4.097 and 5.328, all lie in $\left(\frac\pi4,\frac{7\pi}4\right)$.

Answer: 4