If `3^(sin2x + 2cos^2x) + 3^(1-sin2x + 2sin^2x) = 28` , then the values of x are given by

Options

A. `tan x = 1`
B. `tan x =-1`
C. `cos x = 0`
D. none of these

Answer

To solve the equation \(3^{\sin 2x + 2\cos^2 x} + 3^{1 - \sin 2x + 2\sin^2 x} = 28\), we will follow these steps:

### Step 1: Simplify the Equation
We start with the given equation:
\[ 3^{\sin 2x + 2\cos^2 x} + 3^{1 - \sin 2x + 2\sin^2 x} = 28 \]

### Step 2: Substitute \( \sin^2 x \) and \( \cos^2 x \)
Recall that \( \sin^2 x + \cos^2 x = 1 \). Thus, we can express \( \sin^2 x \) in terms of \( \cos^2 x \):
\[ \sin^2 x = 1 - \cos^2 x \]
Substituting this into the equation:
\[ 3^{\sin 2x + 2\cos^2 x} + 3^{1 - \sin 2x + 2(1 - \cos^2 x)} = 28 \]
This simplifies to:
\[ 3^{\sin 2x + 2\cos^2 x} + 3^{1 - \sin 2x + 2 - 2\cos^2 x} = 28 \]
\[ 3^{\sin 2x + 2\cos^2 x} + 3^{3 - \sin 2x - 2\cos^2 x} = 28 \]

### Step 3: Let \( y = 3^{\sin 2x + 2\cos^2 x} \)
Now, we can rewrite the equation as:
\[ y + \frac{27}{y} = 28 \]
Multiplying through by \( y \) gives:
\[ y^2 + 27 = 28y \]
Rearranging this gives us a quadratic equation:
\[ y^2 - 28y + 27 = 0 \]

### Step 4: Solve the Quadratic Equation
Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ y = \frac{28 \pm \sqrt{28^2 - 4 \cdot 1 \cdot 27}}{2 \cdot 1} \]
Calculating the discriminant:
\[ 28^2 - 4 \cdot 27 = 784 - 108 = 676 \]
Thus:
\[ y = \frac{28 \pm 26}{2} \]
Calculating the two possible values for \( y \):
1. \( y = \frac{54}{2} = 27 \)
2. \( y = \frac{2}{2} = 1 \)

### Step 5: Find Values of \( x \)
Now we have two cases for \( y \):
1. **Case 1**: \( y = 27 \)
\[ 3^{\sin 2x + 2\cos^2 x} = 27 \implies \sin 2x + 2\cos^2 x = 3 \]
Since \( \sin 2x \) can only take values between -1 and 1, this case is not possible.

2. **Case 2**: \( y = 1 \)
\[ 3^{\sin 2x + 2\cos^2 x} = 1 \implies \sin 2x + 2\cos^2 x = 0 \]
This implies:
\[ \sin 2x = -2\cos^2 x \]
Using \( \sin 2x = 2\sin x \cos x \):
\[ 2\sin x \cos x = -2\cos^2 x \]
Dividing by 2 and rearranging gives:
\[ \sin x + \cos x = 0 \implies \tan x = -1 \]

### Step 6: Solve for \( x \)
The general solution for \( \tan x = -1 \) is:
\[ x = \frac{3\pi}{4} + n\pi, \quad n \in \mathbb{Z} \]

### Final Answer
The values of \( x \) are:
\[ x = \frac{3\pi}{4} + n\pi, \quad n \in \mathbb{Z} \]