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 start by noting a key identity: \(\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}\). Let's assign variables for simplicity:

• Let \(a = \sin^{-1} x\) and \(b = \cos^{-1} x\).

According to the identity, \(a + b = \frac{\pi}{2}\).

The given equation is:

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

Substitute \(b = \frac{\pi}{2} - a\) from the identity:

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

Expanding this, we get:

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

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

Rearranging gives:

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

Simplify to find:

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

Thus:

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

Now, since \(a = \sin^{-1} x\), it must be true that \(-\frac{\pi}{2} \leq a \leq \frac{\pi}{2}\). However, \(\frac{11\pi}{10}\) does not fit this range. Therefore, there are no possible values of \(x\) satisfying the original equation within the range for inverse sine.

Given this, the number of real solutions is:

0

Answer 2

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