If `"sin"^(-1)(x^2-7x+12)=npi`, `AAn epsilon I`, then x=

Options

A. -2
B. 4
C. -3
D. 5

Answer

To solve the equation \( \sin^{-1}(x^2 - 7x + 12) = n\pi \) for \( n \in \mathbb{I} \), we follow these steps:

### Step 1: Understand the implications of the equation
The equation \( \sin^{-1}(y) = n\pi \) implies that \( y = \sin(n\pi) \). Since \( \sin(n\pi) = 0 \) for all integers \( n \), we can set the argument of the sine inverse function to zero.

### Step 2: Set the argument to zero
We have:
\[ x^2 - 7x + 12 = 0 \]

### Step 3: Factor the quadratic equation
To solve the quadratic equation, we can factor it. We look for two numbers that multiply to \( 12 \) (the constant term) and add up to \( -7 \) (the coefficient of \( x \)). The numbers \( -3 \) and \( -4 \) fit this requirement.

Thus, we can factor the equation as:
\[ (x - 3)(x - 4) = 0 \]

### Step 4: Solve for \( x \)
Setting each factor equal to zero gives us the solutions:
\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]
\[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \]

### Conclusion
The values of \( x \) that satisfy the original equation are \( x = 3 \) and \( x = 4 \). Since the options provided include \( 4 \), we can conclude that the correct answer is:
\[ \text{The value of } x \text{ is } 4. \]