Solve `2 cos^(-1) x + sin^(-1) x = (2pi)/(3)`

Explanation

To solve the equation \( 2 \cos^{-1} x + \sin^{-1} x = \frac{2\pi}{3} \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 2 \cos^{-1} x + \sin^{-1} x = \frac{2\pi}{3} \]

### Step 2: Use the identity for inverse trigonometric functions
We know that:
\[ \cos^{-1} x + \sin^{-1} x = \frac{\pi}{2} \]
Using this identity, we can rewrite \( 2 \cos^{-1} x \) as:
\[ 2 \cos^{-1} x = \frac{2\pi}{3} - \sin^{-1} x \]
This implies:
\[ \cos^{-1} x + \cos^{-1} x + \sin^{-1} x = \frac{2\pi}{3} \]

### Step 3: Substitute the identity into the equation
Substituting the identity into the equation gives:
\[ \cos^{-1} x + \left(\frac{\pi}{2} - \cos^{-1} x\right) = \frac{2\pi}{3} \]
This simplifies to:
\[ \frac{\pi}{2} + \cos^{-1} x = \frac{2\pi}{3} \]

### Step 4: Isolate \( \cos^{-1} x \)
Now, we isolate \( \cos^{-1} x \):
\[ \cos^{-1} x = \frac{2\pi}{3} - \frac{\pi}{2} \]
To combine these fractions, we find a common denominator:
\[ \cos^{-1} x = \frac{4\pi}{6} - \frac{3\pi}{6} = \frac{\pi}{6} \]

### Step 5: Solve for \( x \)
Now we take the cosine of both sides:
\[ x = \cos\left(\frac{\pi}{6}\right) \]
We know that:
\[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]
Thus, we have:
\[ x = \frac{\sqrt{3}}{2} \]

### Final Answer
The solution to the equation \( 2 \cos^{-1} x + \sin^{-1} x = \frac{2\pi}{3} \) is:
\[ \boxed{\frac{\sqrt{3}}{2}} \]