Question

Let the inverse trigonometric functions take principal values. The number of real solutions of the equation $2 \sin^{-1} x + 3 \cos^{-1} x = \frac{2\pi}{5},$ is ______ .

Answer 1

We are given the equation:

$2 \sin^{-1} x + 3 \cos^{-1} x = \frac{2\pi}{5}$

Let $\sin^{-1} x = \alpha$ and $\cos^{-1} x = \beta$. We know the identity:

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

So, we have:

$2\alpha + 3\beta = \frac{2\pi}{5}$

Using $\beta = \frac{\pi}{2} - \alpha$, substitute this into the equation:

$2\alpha + 3\left(\frac{\pi}{2} - \alpha\right) = \frac{2\pi}{5}$

Simplifying:

$2\alpha + \frac{3\pi}{2} - 3\alpha = \frac{2\pi}{5}$

$-\alpha + \frac{3\pi}{2} = \frac{2\pi}{5}$

$-\alpha = \frac{2\pi}{5} - \frac{3\pi}{2}$

$-\alpha = \frac{4\pi}{10} - \frac{15\pi}{10} = -\frac{11\pi}{10}$

$\alpha = \frac{11\pi}{10}$

Now, since $\alpha = \sin^{-1} x$ and $\sin^{-1} x$ must lie in the range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, we find that $\alpha = \frac{11\pi}{10}$ is not possible because it is outside the allowed range of the inverse sine function.

Thus, the equation has no real solutions.

Final Answer:

0