If `x ne (n pi)/(2), n in Z " and "("cos"x)^("sin"^2 x-3 "sin" x + 2) = 1 " Then," x =`

Options

A. `2n pi + (pi)/(2), n in Z`
B. `(2n + 1) pi - (pi)/(2) , n in Z`
C. `n pi + (-1)^(n) (pi)/(2), n inZ`
D. none of these

Explanation

To solve the equation \( (\cos x)^{(\sin^2 x - 3 \sin x + 2)} = 1 \), we will follow these steps:

### Step 1: Rewrite the equation
We know that any number raised to the power of 0 is equal to 1. Therefore, we can rewrite the equation as:

\[ (\cos x)^{(\sin^2 x - 3 \sin x + 2)} = (\cos x)^0 \]

### Step 2: Set the exponents equal
Since the bases are the same (assuming \(\cos x \neq 0\)), we can set the exponents equal to each other:

\[ \sin^2 x - 3 \sin x + 2 = 0 \]

### Step 3: Factor the quadratic equation
Next, we will factor the quadratic equation:

\[ \sin^2 x - 3 \sin x + 2 = (\sin x - 1)(\sin x - 2) = 0 \]

### Step 4: Solve for \(\sin x\)
Setting each factor to zero gives us:

1. \(\sin x - 1 = 0 \Rightarrow \sin x = 1\)
2. \(\sin x - 2 = 0 \Rightarrow \sin x = 2\)

### Step 5: Analyze the solutions
The sine function has a range of \([-1, 1]\). Therefore, \(\sin x = 2\) is not a valid solution.

The only valid solution is:

\[ \sin x = 1 \]

### Step 6: Find the general solution for \(x\)
The general solution for \(\sin x = 1\) is:

\[ x = \frac{\pi}{2} + 2n\pi \quad \text{for } n \in \mathbb{Z} \]

### Step 7: Consider the restriction
Since the problem states \(x \neq \frac{n\pi}{2}\) for \(n \in \mathbb{Z}\), we need to ensure that our solution does not fall into this category. However, \(\frac{\pi}{2}\) is indeed of the form \(\frac{n\pi}{2}\) where \(n = 1\). Therefore, we need to exclude this solution.

### Final Answer
Thus, the solution to the equation is:

\[ x = n\pi + (-1)^n \frac{\pi}{2} \quad \text{for } n \in \mathbb{Z} \]