Number of solution(s) of the equation `2tan^(-1)(2x-1)=cos^(-1)(x)` is :

Options

A. 1
B. 2
C. 3
D. infinitely many

Answer

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

### Step 1: Define the Functions
Let:
- \( b = \tan^{-1}(2x - 1) \)
- \( a = \cos^{-1}(x) \)

Then, we can rewrite the equation as:
\[ 2b = a \]

### Step 2: Express \( x \) in Terms of \( a \) and \( b \)
From the definitions:
- \( \tan b = 2x - 1 \) implies \( 2x = \tan b + 1 \) or \( x = \frac{\tan b + 1}{2} \)
- \( \cos a = x \)

### Step 3: Use Trigonometric Identities
Using the double angle formula for cosine:
\[ \cos(2b) = \cos^2(b) - \sin^2(b) \]
We can express \( \cos(2b) \) in terms of \( x \):
\[ \cos(2b) = \cos a \]

### Step 4: Express \( \cos(b) \) and \( \sin(b) \)
From the triangle formed by \( b \):
- Opposite side = \( 2x - 1 \)
- Adjacent side = \( 1 \)

Thus, we have:
\[ \cos b = \frac{1}{\sqrt{(2x - 1)^2 + 1}} \]
\[ \sin b = \frac{2x - 1}{\sqrt{(2x - 1)^2 + 1}} \]

### Step 5: Substitute into the Cosine Identity
Substituting into the cosine identity:
\[ \cos(2b) = \frac{1^2 - (2x - 1)^2}{(2x - 1)^2 + 1} \]
This gives:
\[ \cos(2b) = \frac{1 - (4x^2 - 4x + 1)}{(2x - 1)^2 + 1} = \frac{4x - 4x^2}{(2x - 1)^2 + 1} \]

### Step 6: Set Up the Equation
Now we set this equal to \( \cos a = x \):
\[ \frac{4x - 4x^2}{(2x - 1)^2 + 1} = x \]

### Step 7: Clear the Denominator
Multiply both sides by \((2x - 1)^2 + 1\):
\[ 4x - 4x^2 = x((2x - 1)^2 + 1) \]

### Step 8: Expand and Rearrange
Expanding the right side:
\[ 4x - 4x^2 = x(4x^2 - 4x + 1 + 1) \]
\[ 4x - 4x^2 = 4x^3 - 4x^2 + 2x \]

Rearranging gives:
\[ 0 = 4x^3 - 4x^2 + 2x - 4x + 4x^2 \]
\[ 0 = 4x^3 - 2x \]
Factoring out \( 2x \):
\[ 2x(2x^2 - 1) = 0 \]

### Step 9: Solve for \( x \)
This gives us:
1. \( 2x = 0 \) → \( x = 0 \)
2. \( 2x^2 - 1 = 0 \) → \( x^2 = \frac{1}{2} \) → \( x = \pm \frac{1}{\sqrt{2}} \)

### Step 10: Verify Solutions
Now we check each solution in the original equation:
1. For \( x = 0 \):
- \( 2\tan^{-1}(-1) = -\frac{\pi}{2} \) and \( \cos^{-1}(0) = \frac{\pi}{2} \) → Not a solution.

2. For \( x = \frac{1}{\sqrt{2}} \):
- \( 2\tan^{-1}(0) = 0 \) and \( \cos^{-1}(\frac{1}{\sqrt{2}}) = \frac{\pi}{4} \) → Valid solution.

3. For \( x = -\frac{1}{\sqrt{2}} \):
- \( 2\tan^{-1}(-2) \) and \( \cos^{-1}(-\frac{1}{\sqrt{2}}) \) → Not a solution.

### Conclusion
The only valid solution is \( x = \frac{1}{\sqrt{2}} \). Therefore, the number of solutions to the equation is **1**.

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