Which of the following set of values of x satisfies the equation `2^((2 sin^(2) x-3sinx +1))+2^((2-2 sin^(2)x+3 sin x))=9` ?

Options

A. `x =npi pm (pi)/6 , n in I `
B. `x = n pi pm (pi)/3, n in I `
C. `x = n pi , n in I `
D. `x=2n pi +(pi)/(2), n in I `

Answer

To solve the equation \( 2^{(2 \sin^2 x - 3 \sin x + 1)} + 2^{(2 - 2 \sin^2 x + 3 \sin x)} = 9 \), we will follow a systematic approach.

### Step 1: Simplifying the Equation
Let \( K = 2^{(2 \sin^2 x - 3 \sin x + 1)} \). Then, we can rewrite the second term as follows:

\[ 2^{(2 - 2 \sin^2 x + 3 \sin x)} = 2^{(2 - (2 \sin^2 x - 3 \sin x))} = 2^2 \cdot 2^{-(2 \sin^2 x - 3 \sin x)} = 4 \cdot \frac{1}{K} \]

Thus, the equation becomes:

\[ K + \frac{4}{K} = 9 \]

### Step 2: Multiplying through by \( K \)
To eliminate the fraction, multiply both sides by \( K \):

\[ K^2 + 4 = 9K \]

### Step 3: Rearranging the Equation
Rearranging gives us a standard quadratic equation:

\[ K^2 - 9K + 4 = 0 \]

### Step 4: Solving the Quadratic Equation
We can solve this quadratic equation using the quadratic formula:

\[ K = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1, b = -9, c = 4 \):

\[ K = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{9 \pm \sqrt{81 - 16}}{2} = \frac{9 \pm \sqrt{65}}{2} \]

### Step 5: Finding Values of \( K \)
Thus, we have two potential values for \( K \):

\[ K_1 = \frac{9 + \sqrt{65}}{2}, \quad K_2 = \frac{9 - \sqrt{65}}{2} \]

### Step 6: Setting Up the Equations
Now we need to find \( \sin x \) for both values of \( K \):

1. For \( K_1 \):
\[ 2^{(2 \sin^2 x - 3 \sin x + 1)} = K_1 \]

2. For \( K_2 \):
\[ 2^{(2 \sin^2 x - 3 \sin x + 1)} = K_2 \]

### Step 7: Solving for \( \sin x \)
We will solve for \( \sin x \) in both cases.

1. **For \( K_1 \)**:
\[ 2 \sin^2 x - 3 \sin x + 1 = \log_2\left(\frac{9 + \sqrt{65}}{2}\right) \]

2. **For \( K_2 \)**:
\[ 2 \sin^2 x - 3 \sin x + 1 = \log_2\left(\frac{9 - \sqrt{65}}{2}\right) \]

### Step 8: Finding Valid Solutions
Now we need to find the values of \( x \) that satisfy these equations. We know that \( \sin x \) must be in the range \([-1, 1]\).

- For \( K_1 \) and \( K_2 \), we will check if the logarithmic values yield valid \( \sin x \) values.

### Conclusion
The valid solutions for \( x \) will be derived from the valid values of \( \sin x \) obtained from the above equations. The final solutions will be:

- \( x = n\pi + \frac{\pi}{6} \) or \( x = n\pi - \frac{\pi}{6} \) for \( \sin x = \frac{1}{2} \)
- \( x = n\pi + \frac{5\pi}{6} \) or \( x = n\pi - \frac{5\pi}{6} \) for \( \sin x = -\frac{1}{2} \)
- \( x = n\pi \) for \( \sin x = 1 \)

Where \( n \) is any integer.