Solve the equation `4^(sin 2x+2 cos^2 x)+4^(1-sin 2 x+2 sin^2 x)=65`

Options

A. -1
B. 2
C. `sqrt2`
D. 1

Answer

To solve the equation \( 4^{\sin 2x + 2 \cos^2 x} + 4^{1 - \sin 2x + 2 \sin^2 x} = 65 \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ 4^{\sin 2x + 2 \cos^2 x} + 4^{1 - \sin 2x + 2 \sin^2 x} = 65 \]
We can rewrite the terms using the property of exponents:
\[ 4^{\sin 2x} \cdot 4^{2 \cos^2 x} + 4^{1 - \sin 2x} \cdot 4^{2 \sin^2 x} = 65 \]

### Step 2: Simplify the terms
Notice that \( 4^{2 \cos^2 x} = (4^{\cos^2 x})^2 \) and \( 4^{2 \sin^2 x} = (4^{\sin^2 x})^2 \). Let’s denote:
\[ t = 4^{\sin 2x} \]
Then, we can express the equation as:
\[ t \cdot 4^{2 \cos^2 x} + \frac{4}{t} \cdot 4^{2 \sin^2 x} = 65 \]

### Step 3: Substitute \( t \)
We can substitute \( t = 4^{\sin 2x} \) into the equation:
\[ t \cdot 4^{2 \cos^2 x} + \frac{4}{t} \cdot 4^{2 \sin^2 x} = 65 \]
This can be rewritten as:
\[ t \cdot 4^{2 \cos^2 x} + \frac{4^{1 + 2 \sin^2 x}}{t} = 65 \]

### Step 4: Let \( u = 4^{\cos^2 x} \)
Let:
\[ u = 4^{\cos^2 x} \]
Then, the equation becomes:
\[ t \cdot u^2 + \frac{4}{t} \cdot \frac{4}{u} = 65 \]

### Step 5: Multiply through by \( t \)
Multiply the entire equation by \( t \):
\[ t^2 u^2 + 4 = 65t \]
Rearranging gives us:
\[ t^2 u^2 - 65t + 4 = 0 \]

### Step 6: Solve the quadratic equation
This is a quadratic equation in \( t \). We can use the quadratic formula:
\[ t = \frac{65 \pm \sqrt{65^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} \]
Calculating the discriminant:
\[ 65^2 - 16 = 4225 - 16 = 4209 \]
Thus:
\[ t = \frac{65 \pm \sqrt{4209}}{2} \]

### Step 7: Find values of \( t \)
Calculating \( \sqrt{4209} \) gives approximately \( 64.9 \):
\[ t = \frac{65 \pm 64.9}{2} \]
This gives us two possible values for \( t \):
1. \( t_1 \approx \frac{129.9}{2} \approx 64.95 \)
2. \( t_2 \approx \frac{0.1}{2} \approx 0.05 \)

### Step 8: Solve for \( x \)
Recall that \( t = 4^{\sin 2x} \):
1. For \( t_1 \approx 64 \):
\[ 4^{\sin 2x} = 64 \implies \sin 2x = 3 \]
(not possible since \( \sin \) cannot exceed 1)

2. For \( t_2 \approx 0.05 \):
\[ 4^{\sin 2x} = 0.05 \implies \sin 2x = -1 \]
Thus:
\[ 2x = \frac{3\pi}{2} + 2n\pi \implies x = \frac{3\pi}{4} + n\pi \]

### Final Solution
The solution to the equation is:
\[ x = \frac{3\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]