JEE Main 2024 — 9 April, Shift 2 — Question 29

We are asked to find the number of real solutions of

$$2 \sin^{-1} x + 3 \cos^{-1} x = \frac{2\pi}{5}, \quad x \in [-1,1].$$

Use the relation between $\sin^{-1} x$ and $\cos^{-1} x$:

$$\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \implies \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x.$$

Substitute in the equation:

$$2 \sin^{-1} x + 3\left(\frac{\pi}{2} - \sin^{-1} x\right) = \frac{2\pi}{5}$$ $$2 \sin^{-1} x + \frac{3\pi}{2} - 3 \sin^{-1} x = \frac{2\pi}{5}$$ $$-\sin^{-1} x + \frac{3\pi}{2} = \frac{2\pi}{5}$$

Solve for $\sin^{-1} x$:

$$-\sin^{-1} x = \frac{2\pi}{5} - \frac{3\pi}{2} = -\frac{11\pi}{10} \implies \sin^{-1} x = \frac{11\pi}{10}.$$

Check feasibility:

The principal value of $\sin^{-1} x$ satisfies

$$\sin^{-1} x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].$$

But $\frac{11\pi}{10} > \frac{\pi}{2}$, which is outside the allowed range.

There are no real solutions. Hence, the number of real solutions is

$$\boxed{0}.$$