Find the general solution of the equation `3^(sin2x+2cos^2x)+3^(1-sin2x+2sin^2x)=28`

Options

A. 3
B. 4
C. 5
D. 6

Explanation

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: Rewrite the equation
We start with the given equation:
\[ 3^{\sin 2x + 2\cos^2 x} + 3^{1 - \sin 2x + 2\sin^2 x} = 28 \]
Notice that \( 2\cos^2 x + 2\sin^2 x = 2 \) (since \(\cos^2 x + \sin^2 x = 1\)). Thus, we can rewrite the second exponent:
\[ 1 - \sin 2x + 2\sin^2 x = 1 - \sin 2x + 2(1 - \cos^2 x) = 1 - \sin 2x + 2 - 2\cos^2 x = 3 - \sin 2x - 2\cos^2 x \]

### Step 2: Simplify the equation
Now, we can rewrite the equation as:
\[ 3^{\sin 2x + 2\cos^2 x} + 3^{3 - \sin 2x - 2\cos^2 x} = 28 \]
Let \( y = 3^{\sin 2x + 2\cos^2 x} \). Then, the equation becomes:
\[ y + \frac{27}{y} = 28 \]

### Step 3: Multiply through by \( y \)
To eliminate the fraction, multiply through by \( y \):
\[ y^2 + 27 = 28y \]
Rearranging gives us a quadratic equation:
\[ y^2 - 28y + 27 = 0 \]

### Step 4: Solve the quadratic equation
We can use 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:
\[ 28^2 - 4 \cdot 1 \cdot 27 = 784 - 108 = 676 \]
Thus,
\[ y = \frac{28 \pm \sqrt{676}}{2} = \frac{28 \pm 26}{2} \]
Calculating the two possible values for \( y \):
1. \( y = \frac{54}{2} = 27 \)
2. \( y = \frac{2}{2} = 1 \)

### Step 5: Find \( \sin 2x + 2\cos^2 x \)
Recall \( y = 3^{\sin 2x + 2\cos^2 x} \):
1. If \( y = 27 \), then \( \sin 2x + 2\cos^2 x = 3 \).
2. If \( y = 1 \), then \( \sin 2x + 2\cos^2 x = 0 \).

### Step 6: Solve for \( x \)
**Case 1:** \( \sin 2x + 2\cos^2 x = 3 \)

This case is not possible since \( \sin 2x \) can only take values between -1 and 1, and \( 2\cos^2 x \) can take values between 0 and 2. Thus, \( \sin 2x + 2\cos^2 x \) can only take values between -1 and 3, but cannot equal 3.

**Case 2:** \( \sin 2x + 2\cos^2 x = 0 \)

Using the identity \( \cos^2 x = 1 - \sin^2 x \):
\[ \sin 2x + 2(1 - \sin^2 x) = 0 \]
This simplifies to:
\[ \sin 2x + 2 - 2\sin^2 x = 0 \]
Using \( \sin 2x = 2\sin x \cos x \):
\[ 2\sin x \cos x + 2 - 2\sin^2 x = 0 \]
Dividing by 2 gives:
\[ \sin x \cos x + 1 - \sin^2 x = 0 \]
Rearranging gives:
\[ \sin^2 x - \sin x \cos x - 1 = 0 \]

### Step 7: Solve the quadratic in \( \sin x \)
Let \( t = \sin x \):
\[ t^2 - t\sqrt{1-t^2} - 1 = 0 \]
This is a more complex equation, but we can find solutions for \( x \) using numerical methods or graphing.

### Step 8: General solution
From the equation \( \sin 2x + \cos 2x = -1 \):
This leads to:
\[ 2x = 3\pi/4 + n\pi \quad \text{where } n \in \mathbb{Z} \]
Thus:
\[ x = \frac{3\pi}{8} + \frac{n\pi}{2} \]

### Final Answer
The general solution is:
\[ x = \frac{3\pi}{8} + \frac{n\pi}{2}, \quad n \in \mathbb{Z} \]