JEE Main 2026 — 28 January, Morning Shift — Question 22

Let $\theta=\frac{1}{2} \sin ^{-1} \frac{2}{3}$, then $\frac{1}{2} \cos ^{-1} \frac{1}{3}=\left(\frac{\pi}{4}-\theta\right)$

$\mathrm{k}=\tan \theta+\cot \theta=\frac{1}{\sin \theta \cos \theta}=\frac{2}{\sin 2 \theta}$

$\mathrm{k}=\frac{2}{\frac{2}{3}}=3$

$\sin ^{-1}(3 \mathrm{x}-1)=\sin ^{-1} \mathrm{x}-\cos ^{-1} \mathrm{x}$

$\sin ^{-1}(3 \mathrm{x}-1)=\frac{\pi}{2}-2 \cos ^{-1} \mathrm{x}$

$3 \mathrm{x}-1=\sin \left(\frac{\pi}{2}-2 \cos ^{-1} \mathrm{x}\right)$

$3 \mathrm{x}-1=2 \mathrm{x}^{2}-1$

$\Rightarrow \mathrm{x}=0, \frac{3}{2}$ (rejected)

No. of solution $=1$.