Number of solutions of the equation
`2(sin^(-1)x)^2-sin^(-1)x-6=0` is

Options

A. 2
B. 1
C. 0
D. 3

Answer

To solve the equation \(2(\sin^{-1}x)^2 - \sin^{-1}x - 6 = 0\) and find the number of solutions, we can follow these steps:

### Step 1: Substitution
Let \(y = \sin^{-1}x\). Then, the equation becomes:
\[ 2y^2 - y - 6 = 0 \]

### Step 2: Rearranging the Equation
We can rearrange the equation as follows:
\[ 2y^2 - y - 6 = 0 \]

### Step 3: Factoring the Quadratic Equation
Next, we will factor the quadratic equation. We need to find two numbers that multiply to \(2 \times (-6) = -12\) and add to \(-1\). The numbers \(-4\) and \(3\) fit this requirement. Thus, we can rewrite the equation as:
\[ 2y^2 - 4y + 3y - 6 = 0 \]
Grouping the terms gives:
\[ 2y(y - 2) + 3(y - 2) = 0 \]
Factoring out the common term:
\[ (2y + 3)(y - 2) = 0 \]

### Step 4: Finding the Values of \(y\)
Setting each factor to zero gives us:
1. \(2y + 3 = 0 \implies y = -\frac{3}{2}\)
2. \(y - 2 = 0 \implies y = 2\)

### Step 5: Validating the Solutions
Now, we need to check if these values of \(y\) are valid in the context of the inverse sine function:
- The range of \(\sin^{-1}x\) is \([- \frac{\pi}{2}, \frac{\pi}{2}]\) or approximately \([-1.57, 1.57]\).

Checking the values:
- \(y = -\frac{3}{2} \approx -1.5\) is within the valid range.
- \(y = 2\) is outside the valid range.

### Conclusion
Since only \(y = -\frac{3}{2}\) is valid, we conclude that there is only **one solution** for the equation \(2(\sin^{-1}x)^2 - \sin^{-1}x - 6 = 0\).

Thus, the number of solutions is:
\[ \boxed{1} \]