If `cos(2sin^(-1)x)=1/9 , " then " x=`

Options

A. `2//3`
B. `-2//3`
C. `1//3`
D. `-1//3`

Answer

To solve the equation \( \cos(2 \sin^{-1} x) = \frac{1}{9} \), we can follow these steps:

### Step 1: Set up the equation
Let \( \theta = \sin^{-1} x \). Then, we have:
\[ \cos(2\theta) = \frac{1}{9} \]

### Step 2: Use the double angle formula for cosine
The double angle formula for cosine states:
\[ \cos(2\theta) = 1 - 2\sin^2(\theta) \]
Substituting this into our equation gives:
\[ 1 - 2\sin^2(\theta) = \frac{1}{9} \]

### Step 3: Rearrange the equation
Rearranging the equation, we get:
\[ 2\sin^2(\theta) = 1 - \frac{1}{9} \]
Calculating the right-hand side:
\[ 1 - \frac{1}{9} = \frac{9}{9} - \frac{1}{9} = \frac{8}{9} \]
Thus, we have:
\[ 2\sin^2(\theta) = \frac{8}{9} \]

### Step 4: Solve for \(\sin^2(\theta)\)
Dividing both sides by 2:
\[ \sin^2(\theta) = \frac{8}{9} \cdot \frac{1}{2} = \frac{8}{18} = \frac{4}{9} \]

### Step 5: Take the square root
Taking the square root of both sides gives:
\[ \sin(\theta) = \pm \frac{2}{3} \]

### Step 6: Relate back to \(x\)
Since we defined \( \theta = \sin^{-1} x \), we have:
\[ x = \sin(\theta) = \pm \frac{2}{3} \]

### Step 7: Final answer
Thus, the possible values of \( x \) are:
\[ x = \frac{2}{3} \quad \text{or} \quad x = -\frac{2}{3} \]

### Summary
The final values of \( x \) are \( \frac{2}{3} \) and \( -\frac{2}{3} \).

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