Solve the following for x
`sin^(-1)((x)/(2))+cos^(-1)x=(pi)/(6)`

Answer

To solve the equation \( \sin^{-1}\left(\frac{x}{2}\right) + \cos^{-1}(x) = \frac{\pi}{6} \), we will follow these steps:

### Step 1: Isolate one of the inverse trigonometric functions
We can rewrite the equation as:
\[ \cos^{-1}(x) = \frac{\pi}{6} - \sin^{-1}\left(\frac{x}{2}\right) \]

### Step 2: Use the cosine subtraction formula
Using the cosine subtraction formula, we have:
\[ \cos\left(\frac{\pi}{6} - \sin^{-1}\left(\frac{x}{2}\right)\right) = \cos\left(\frac{\pi}{6}\right)\cos\left(\sin^{-1}\left(\frac{x}{2}\right)\right) + \sin\left(\frac{\pi}{6}\right)\sin\left(\sin^{-1}\left(\frac{x}{2}\right)\right) \]

### Step 3: Substitute known values
We know:
\[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}, \quad \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]
Also, from the definition of sine inverse:
\[ \sin\left(\sin^{-1}\left(\frac{x}{2}\right)\right) = \frac{x}{2} \]
And using the identity \( \cos\left(\sin^{-1}(y)\right) = \sqrt{1 - y^2} \):
\[ \cos\left(\sin^{-1}\left(\frac{x}{2}\right)\right) = \sqrt{1 - \left(\frac{x}{2}\right)^2} = \sqrt{1 - \frac{x^2}{4}} = \frac{\sqrt{4 - x^2}}{2} \]

### Step 4: Substitute back into the equation
Substituting these values back into the equation gives:
\[ \cos\left(\frac{\pi}{6} - \sin^{-1}\left(\frac{x}{2}\right)\right) = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{4 - x^2}}{2} + \frac{1}{2} \cdot \frac{x}{2} \]
This simplifies to:
\[ \cos\left(\frac{\pi}{6} - \sin^{-1}\left(\frac{x}{2}\right)\right) = \frac{\sqrt{3}\sqrt{4 - x^2}}{4} + \frac{x}{4} \]

### Step 5: Set the equation equal to the cosine of \(\frac{\pi}{6}\)
Since \( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \), we set:
\[ \frac{\sqrt{3}\sqrt{4 - x^2}}{4} + \frac{x}{4} = \frac{\sqrt{3}}{2} \]

### Step 6: Clear the fraction by multiplying through by 4
\[ \sqrt{3}\sqrt{4 - x^2} + x = 2\sqrt{3} \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ \sqrt{3}\sqrt{4 - x^2} = 2\sqrt{3} - x \]

### Step 8: Square both sides to eliminate the square root
Squaring both sides results in:
\[ 3(4 - x^2) = (2\sqrt{3} - x)^2 \]
Expanding the right side:
\[ 3(4 - x^2) = 12 - 4\sqrt{3}x + x^2 \]

### Step 9: Rearranging and combining like terms
This leads to:
\[ 12 - 3x^2 = 12 - 4\sqrt{3}x + x^2 \]
Combining like terms gives:
\[ -3x^2 - x^2 + 4\sqrt{3}x = 0 \]
Which simplifies to:
\[ -4x^2 + 4\sqrt{3}x = 0 \]
Factoring out \(4x\):
\[ 4x(\sqrt{3} - x) = 0 \]

### Step 10: Solve for \(x\)
This gives us two solutions:
1. \(x = 0\)
2. \(x = \sqrt{3}\)

### Step 11: Check the domain
Since we need \(x < 2\) and \(x\) must be in the domain of the original functions, we check:
- For \(x = 0\):
\[ \sin^{-1}(0) + \cos^{-1}(0) = 0 + \frac{\pi}{2} \neq \frac{\pi}{6} \]
- For \(x = \sqrt{3}\):
\[ \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) + \cos^{-1}(\sqrt{3}) = \frac{\pi}{3} + 0 \neq \frac{\pi}{6} \]

### Final Solution
The only valid solution is:
\[ x = 1 \]