Solve `2^(cos 2x)+1=3.2^(-sin^(2) x)`

Explanation

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

### Step 1: Rewrite the equation
Start with the original equation:
\[ 2^{\cos 2x} + 1 = 3 \cdot 2^{-\sin^2 x} \]

### Step 2: Substitute for \(\cos 2x\)
Using the identity \(\cos 2x = 1 - 2\sin^2 x\), we can substitute:
\[ 2^{1 - 2\sin^2 x} + 1 = 3 \cdot 2^{-\sin^2 x} \]

### Step 3: Rewrite the left side
We can express the left side as:
\[ 2^1 \cdot 2^{-2\sin^2 x} + 1 = 3 \cdot 2^{-\sin^2 x} \]
This simplifies to:
\[ 2 \cdot 2^{-2\sin^2 x} + 1 = 3 \cdot 2^{-\sin^2 x} \]

### Step 4: Let \( t = \sin^2 x \)
Let \( t = \sin^2 x \). Then we can rewrite the equation:
\[ 2 \cdot 2^{-2t} + 1 = 3 \cdot 2^{-t} \]

### Step 5: Simplify the equation
This can be rewritten as:
\[ 2 \cdot \frac{1}{2^{2t}} + 1 = 3 \cdot \frac{1}{2^t} \]
Multiplying through by \( 2^{2t} \) to eliminate the fractions gives:
\[ 2 + 2^{2t} = 3 \cdot 2^t \]

### Step 6: Rearrange into a standard quadratic form
Rearranging the equation, we have:
\[ 2^{2t} - 3 \cdot 2^t + 2 = 0 \]
Let \( u = 2^t \). Then the equation becomes:
\[ u^2 - 3u + 2 = 0 \]

### Step 7: Factor the quadratic equation
Factoring gives:
\[ (u - 1)(u - 2) = 0 \]
Thus, we have:
\[ u = 1 \quad \text{or} \quad u = 2 \]

### Step 8: Solve for \( t \)
Substituting back for \( u \):
1. If \( u = 1 \):
\[ 2^t = 1 \implies t = 0 \implies \sin^2 x = 0 \implies \sin x = 0 \implies x = n\pi \]
2. If \( u = 2 \):
\[ 2^t = 2 \implies t = 1 \implies \sin^2 x = 1 \implies \sin x = \pm 1 \implies x = n\pi \pm \frac{\pi}{2} \]

### Final Solutions
Thus, the solutions to the original equation are:
\[ x = n\pi \quad \text{and} \quad x = n\pi \pm \frac{\pi}{2} \]