If `6 sin^(-1)(x^(2)-6x+17/2)=pi` then

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the equation \( 6 \sin^{-1}\left(x^2 - 6x + \frac{17}{2}\right) = \pi \), we will follow these steps:

### Step 1: Isolate the inverse sine function
We start by dividing both sides of the equation by 6:
\[ \sin^{-1}\left(x^2 - 6x + \frac{17}{2}\right) = \frac{\pi}{6} \]

### Step 2: Apply the sine function
Next, we apply the sine function to both sides to eliminate the inverse sine:
\[ x^2 - 6x + \frac{17}{2} = \sin\left(\frac{\pi}{6}\right) \]
Since \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\), we can rewrite the equation as:
\[ x^2 - 6x + \frac{17}{2} = \frac{1}{2} \]

### Step 3: Rearrange the equation
Now, we will move \(\frac{1}{2}\) to the left side:
\[ x^2 - 6x + \frac{17}{2} - \frac{1}{2} = 0 \]
This simplifies to:
\[ x^2 - 6x + \frac{16}{2} = 0 \]
or
\[ x^2 - 6x + 8 = 0 \]

### Step 4: Factor the quadratic equation
Now we need to factor the quadratic equation \(x^2 - 6x + 8 = 0\). We look for two numbers that multiply to 8 and add up to -6. The numbers -2 and -4 work:
\[ (x - 2)(x - 4) = 0 \]

### Step 5: Solve for \(x\)
Setting each factor equal to zero gives us the solutions:
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]
\[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \]

### Conclusion
The solutions to the equation are \(x = 2\) and \(x = 4\).