The solution of the equation `cos^(-1)x+cos^(-1)2x=(2pi)/(3)` is

Options

A. `x=pm(sqrt3)/(sqrt28)`
B. `x =pm(sqrt3)/(sqrt26)`
C. `x=pm(1)/(sqrt2)`
D. `x=pm(sqrt3) /(sqrt38) `

Answer

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

### Step 1: Use the formula for the sum of inverse cosines
We know that:
\[ \cos^{-1} a + \cos^{-1} b = \cos^{-1(ab - \sqrt{(1-a^2)(1-b^2)})} \]
In our case, let \( a = x \) and \( b = 2x \). Then we can rewrite the equation as:
\[ \cos^{-1} x + \cos^{-1} 2x = \cos^{-1(x \cdot 2x - \sqrt{(1-x^2)(1-(2x)^2)})} \]
This simplifies to:
\[ \cos^{-1} x + \cos^{-1} 2x = \cos^{-1(2x^2 - \sqrt{(1-x^2)(1-4x^2)})} \]

### Step 2: Set the cosine of both sides
Since \( \cos^{-1}(y) = \theta \) implies \( \cos(\theta) = y \), we can set:
\[ 2x^2 - \sqrt{(1-x^2)(1-4x^2)} = \cos\left(\frac{2\pi}{3}\right) \]
We know that \( \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \). Thus, we have:
\[ 2x^2 - \sqrt{(1-x^2)(1-4x^2)} = -\frac{1}{2} \]

### Step 3: Rearranging the equation
Rearranging gives us:
\[ 2x^2 + \frac{1}{2} = \sqrt{(1-x^2)(1-4x^2)} \]

### Step 4: Square both sides
Now, we square both sides to eliminate the square root:
\[ \left(2x^2 + \frac{1}{2}\right)^2 = (1-x^2)(1-4x^2) \]
Expanding both sides:
\[ 4x^4 + 2x^2 + \frac{1}{4} = 1 - 4x^2 - x^2 + 4x^4 \]
This simplifies to:
\[ 4x^4 + 2x^2 + \frac{1}{4} = 1 - 5x^2 + 4x^4 \]

### Step 5: Move all terms to one side
Now, we can move all terms to one side:
\[ 4x^4 - 4x^4 + 2x^2 + 5x^2 + \frac{1}{4} - 1 = 0 \]
This simplifies to:
\[ 7x^2 - \frac{3}{4} = 0 \]

### Step 6: Solve for \( x^2 \)
Rearranging gives us:
\[ 7x^2 = \frac{3}{4} \]
\[ x^2 = \frac{3}{28} \]

### Step 7: Solve for \( x \)
Taking the square root:
\[ x = \pm \sqrt{\frac{3}{28}} = \pm \frac{\sqrt{3}}{\sqrt{28}} = \pm \frac{\sqrt{3}}{2\sqrt{7}} = \pm \frac{\sqrt{21}}{14} \]

### Final Answer
Thus, the solution to the equation is:
\[ x = \pm \frac{\sqrt{3}}{2\sqrt{7}} \]