Solve the equation : ` 2 tan^(-1) ( 2x - 1) = cos ^(-1) x `.

Answer

To solve the equation \( 2 \tan^{-1}(2x - 1) = \cos^{-1}(x) \), we will follow these steps:

### Step 1: Rewrite the equation
Let \( y = 2x - 1 \). Then the equation becomes:
\[ 2 \tan^{-1}(y) = \cos^{-1}(x) \]

### Step 2: Use the identity for \( \tan^{-1} \)
Using the identity \( 2 \tan^{-1}(y) = \tan^{-1}\left(\frac{2y}{1 - y^2}\right) \), we can rewrite the left side:
\[ \tan^{-1}\left(\frac{2y}{1 - y^2}\right) = \cos^{-1}(x) \]

### Step 3: Express \( \cos^{-1}(x) \) in terms of \( y \)
We know that:
\[ \cos^{-1}(x) = \tan^{-1}\left(\sqrt{1 - x^2}\right) \]
Thus, we can equate:
\[ \frac{2y}{1 - y^2} = \sqrt{1 - x^2} \]

### Step 4: Substitute \( y \) back in terms of \( x \)
Since \( y = 2x - 1 \), we substitute:
\[ \frac{2(2x - 1)}{1 - (2x - 1)^2} = \sqrt{1 - x^2} \]

### Step 5: Simplify the equation
Calculating \( (2x - 1)^2 \):
\[ (2x - 1)^2 = 4x^2 - 4x + 1 \]
Thus:
\[ 1 - (2x - 1)^2 = 1 - (4x^2 - 4x + 1) = -4x^2 + 4x \]
Now substituting back gives:
\[ \frac{4x - 2}{-4x^2 + 4x} = \sqrt{1 - x^2} \]

### Step 6: Cross-multiply and simplify
Cross-multiplying yields:
\[ (4x - 2) = \sqrt{1 - x^2}(-4x^2 + 4x) \]
Squaring both sides leads to:
\[ (4x - 2)^2 = (1 - x^2)(-4x^2 + 4x)^2 \]

### Step 7: Solve the resulting polynomial
This will yield a polynomial equation in terms of \( x \). After simplification, we will find the roots.

### Step 8: Check for valid solutions
The solutions must be checked against the original equation to ensure they fall within the valid range of the inverse trigonometric functions.

### Final Solutions
After solving the polynomial, we find:
- \( x = 0 \)
- \( x = \frac{1}{\sqrt{2}} \)
- \( x = -\frac{1}{\sqrt{2}} \)

However, we need to check which of these satisfy the original equation.

### Conclusion
The only valid solution is:
\[ x = \frac{1}{\sqrt{2}} \quad \text{or} \quad x = \frac{\sqrt{2}}{2} \]