The given equation is:

$\cos^{-1}(x) + \sin^{-1}(x) = \pi$

We know that there is an important identity involving inverse trigonometric functions:

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

However, the equation here states that the sum is equal to $\pi$, which is not possible for real values of $x$. This is because:

$\cos^{-1}(x) + \sin^{-1}(x) = \frac{\pi}{2} \quad \text{for all} \quad x \in [-1, 1]$

Thus, the equation $\cos^{-1}(x) + \sin^{-1}(x) = \pi$ does not hold for any real value of $x$.

Conclusion:

There is no real solution to the given equation.

Would you like more details or have any questions?

Here are 5 related questions:

1. What are the properties of inverse trigonometric functions?
2. Can inverse trigonometric functions be used to solve real-world problems?
3. How do you derive the identity $\cos^{-1}(x) + \sin^{-1}(x) = \frac{\pi}{2}$?
4. What are the domains and ranges of $\cos^{-1}(x)$ and $\sin^{-1}(x)$?
5. What happens when we take the inverse of other trigonometric functions?

Tip: Always verify whether the identity used is valid for the equation you are solving!