The equation provided is:

\[
\cos^{-1}(x) + \sin^{-1}\left(\frac{x}{2}\right) = \pi
\]

### Step-by-step solution:

1. **Using known identity:**
   The identity \( \cos^{-1}(x) + \sin^{-1}(x) = \frac{\pi}{2} \) holds for any \( x \in [-1, 1] \). However, the equation here sums to \( \pi \), which suggests this is not a standard case. Let’s explore if we can still solve it.

2. **Rearranging the equation:**
   First, isolate one of the inverse trigonometric terms:
   \[
   \cos^{-1}(x) = \pi - \sin^{-1}\left(\frac{x}{2}\right)
   \]

3. **Taking cosine on both sides:**
   Apply the cosine function to both sides:
   \[
   x = \cos\left(\pi - \sin^{-1}\left(\frac{x}{2}\right)\right)
   \]
   Using the trigonometric identity \( \cos(\pi - \theta) = -\cos(\theta) \), this simplifies to:
   \[
   x = -\cos\left(\sin^{-1}\left(\frac{x}{2}\right)\right)
   \]

4. **Simplifying the cosine of an inverse sine:**
   We know that \( \cos(\sin^{-1}(y)) = \sqrt{1 - y^2} \). So, for \( y = \frac{x}{2} \):
   \[
   \cos\left(\sin^{-1}\left(\frac{x}{2}\right)\right) = \sqrt{1 - \left(\frac{x}{2}\right)^2} = \sqrt{1 - \frac{x^2}{4}}
   \]
   Thus, the equation becomes:
   \[
   x = -\sqrt{1 - \frac{x^2}{4}}
   \]

5. **Squaring both sides:**
   To remove the square root, square both sides:
   \[
   x^2 = 1 - \frac{x^2}{4}
   \]

6. **Solving the quadratic equation:**
   Multiply through by 4 to eliminate the fraction:
   \[
   4x^2 = 4 - x^2
   \]
   Combine like terms:
   \[
   5x^2 = 4
   \]
   Divide by 5:
   \[
   x^2 = \frac{4}{5}
   \]
   Taking the square root of both sides:
   \[
   x = \pm \frac{2}{\sqrt{5}} = \pm \frac{2\sqrt{5}}{5}
   \]

### Conclusion:
The solution is:
\[
x = \pm \frac{2\sqrt{5}}{5}
\]

Would you like more details or have any questions?

Here are 5 related questions:
1. How do we handle trigonometric identities when inverse functions are involved?
2. Why does squaring both sides require careful checking of solutions?
3. Can the identity \( \cos(\pi - \theta) = -\cos(\theta) \) be used for all values of \( \theta \)?
4. What are the applications of inverse trigonometric equations in physics?
5. How do you find the domain and range of inverse sine and cosine functions?

**Tip:** Always check for extraneous solutions when squaring both sides of an equation!

Learn how to solve the equation \( \cos^{-1}(x) + \sin^{-1}\left(\frac{x}{2}\right) = \pi \) step-by-step. This problem involves trigonometric identities and inverse trigonometric functions, suitable for advanced high school students or those studying trigonometry.

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