The equation `sin^(-1)x - 3sin^(-1)a =0` has real solutions for x if

Options

A. `a in R`
B. `a in[-1, 1]`
C. `a in [0, 1/2]`
D. `a in [-1/2, 1/2]`

Answer

To solve the equation \( \sin^{-1}x - 3\sin^{-1}a = 0 \) for real solutions for \( x \), we can follow these steps:

### Step 1: Rearranging the Equation
Start by rearranging the equation:
\[ \sin^{-1}x = 3\sin^{-1}a \]

### Step 2: Understanding the Range of \( \sin^{-1} \)
The function \( \sin^{-1}x \) (inverse sine) has a range of \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \). Therefore, for \( 3\sin^{-1}a \) to also be in this range, we need:
\[ -\frac{\pi}{2} \leq 3\sin^{-1}a \leq \frac{\pi}{2} \]

### Step 3: Dividing by 3
To isolate \( \sin^{-1}a \), divide the entire inequality by 3:
\[ -\frac{\pi}{6} \leq \sin^{-1}a \leq \frac{\pi}{6} \]

### Step 4: Converting to Sine Values
Now, we convert the bounds back to sine values. We know:
\[ \sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2} \quad \text{and} \quad \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]
Thus, we can rewrite the inequality as:
\[ -\frac{1}{2} \leq a \leq \frac{1}{2} \]

### Step 5: Conclusion
The condition for \( a \) to ensure that the equation \( \sin^{-1}x - 3\sin^{-1}a = 0 \) has real solutions for \( x \) is:
\[ a \in \left[-\frac{1}{2}, \frac{1}{2}\right] \]

### Final Answer
The values of \( a \) must lie in the interval:
\[ a \in \left[-\frac{1}{2}, \frac{1}{2}\right] \]

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