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

Correct Answer: 0

To solve the equation \(2 \sin^{-1} x + 3 \cos^{-1} x = \frac{2\pi}{5}\), we utilize the identity \(\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}\). Let \(a = \sin^{-1} x\) and \(b = \cos^{-1} x\). Thus, \(a + b = \frac{\pi}{2}\).

The equation can be written as \(2a + 3b = \frac{2\pi}{5}\).

Substituting \(b = \frac{\pi}{2} - a\) into the equation yields:

\[2a + 3\left(\frac{\pi}{2} - a\right) = \frac{2\pi}{5}\]

Expanding this expression results in:

\[2a + \frac{3\pi}{2} - 3a = \frac{2\pi}{5}\]

\[-a + \frac{3\pi}{2} = \frac{2\pi}{5}\]

Rearranging to isolate \(a\):

\[-a = \frac{2\pi}{5} - \frac{3\pi}{2}\]

Simplifying the right side:

\[-a = \frac{2\pi - 7.5\pi}{5} = -\frac{5.5\pi}{5}\]

Therefore:

\[a = \frac{5.5\pi}{5} = \frac{11\pi}{10}\]

However, the range for \(a = \sin^{-1} x\) is \(-\frac{\pi}{2} \leq a \leq \frac{\pi}{2}\). The calculated value \(a = \frac{11\pi}{10}\) falls outside this valid range.

Consequently, there are no real values of \(x\) that satisfy the original equation within the defined domain for inverse sine.

The number of real solutions is:

0