The general solution of the equation `2^(cos2x)+1=3*2^(-sin^2x)` is

Options

A. `npi, n in I`
B. `npi +pi/2, n in I`
C. `npi - pi/2 in I`
D. `npi + pi/3, n in I`

Answer

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

### Step 1: Rewrite the equation
The given equation is:
\[ 2^{\cos 2x} + 1 = 3 \cdot 2^{-\sin^2 x} \]
We can rewrite \( \cos 2x \) using the identity:
\[ \cos 2x = 1 - 2\sin^2 x \]
Thus, we can substitute this into the equation:
\[ 2^{1 - 2\sin^2 x} + 1 = 3 \cdot 2^{-\sin^2 x} \]

### Step 2: Simplify the left-hand side
Using the property of exponents, we can simplify the left-hand side:
\[ 2^{1 - 2\sin^2 x} = \frac{2}{2^{2\sin^2 x}} \]
So the equation becomes:
\[ \frac{2}{2^{2\sin^2 x}} + 1 = 3 \cdot 2^{-\sin^2 x} \]

### Step 3: Multiply through by \( 2^{2\sin^2 x} \)
To eliminate the fraction, multiply both sides by \( 2^{2\sin^2 x} \):
\[ 2 + 2^{2\sin^2 x} = 3 \cdot 2^{2\sin^2 x - \sin^2 x} = 3 \cdot 2^{\sin^2 x} \]
This simplifies to:
\[ 2 + 2^{2\sin^2 x} = 3 \cdot 2^{\sin^2 x} \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ 2^{2\sin^2 x} - 3 \cdot 2^{\sin^2 x} + 2 = 0 \]
Let \( t = 2^{\sin^2 x} \). Then, we have:
\[ t^2 - 3t + 2 = 0 \]

### Step 5: Solve the quadratic equation
Factoring the quadratic:
\[ (t - 1)(t - 2) = 0 \]
Thus, the solutions for \( t \) are:
\[ t = 1 \quad \text{or} \quad t = 2 \]

### Step 6: Back substitute for \( \sin^2 x \)
Substituting back for \( t \):
1. For \( t = 1 \):
\[ 2^{\sin^2 x} = 1 \implies \sin^2 x = 0 \implies \sin x = 0 \]
This gives:
\[ x = n\pi, \quad n \in \mathbb{Z} \]

2. For \( t = 2 \):
\[ 2^{\sin^2 x} = 2 \implies \sin^2 x = 1 \implies \sin x = \pm 1 \]
This gives:
\[ x = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z} \]

### Final General Solution
Combining both results, the general solution of the equation is:
\[ x = n\pi \quad \text{or} \quad x = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z} \]