Find the most general solution for ` 2^(sin x )+2^( cos x )=2^((1-(1/sqrt2)))`

Answer

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

### Step 1: Rewrite the equation
We can rewrite the right-hand side of the equation as follows:
\[ 2^{\sin x} + 2^{\cos x} = 2^1 \cdot 2^{-\frac{1}{\sqrt{2}}} \]
This simplifies to:
\[ 2^{\sin x} + 2^{\cos x} = 2^{1 - \frac{1}{\sqrt{2}}} \]

### Step 2: Divide both sides by \(2^{-\frac{1}{\sqrt{2}}}\)
We can divide the entire equation by \(2^{-\frac{1}{\sqrt{2}}}\):
\[ \frac{2^{\sin x}}{2^{-\frac{1}{\sqrt{2}}}} + \frac{2^{\cos x}}{2^{-\frac{1}{\sqrt{2}}}} = 2^1 \]
This simplifies to:
\[ 2^{\sin x + \frac{1}{\sqrt{2}}} + 2^{\cos x + \frac{1}{\sqrt{2}}} = 2 \]

### Step 3: Set up equations
Now, we can express this as:
\[ 2^{\sin x + \frac{1}{\sqrt{2}}} + 2^{\cos x + \frac{1}{\sqrt{2}}} = 2^1 \]
Since \(2^1 = 2^0 + 2^0\), we can compare both sides:
\[ \sin x + \frac{1}{\sqrt{2}} = 0 \quad \text{and} \quad \cos x + \frac{1}{\sqrt{2}} = 0 \]

### Step 4: Solve for \(\sin x\)
From the first equation:
\[ \sin x = -\frac{1}{\sqrt{2}} \]
This corresponds to:
\[ x = n\pi - \frac{\pi}{4} \quad \text{where } n \in \mathbb{Z} \]

### Step 5: Solve for \(\cos x\)
From the second equation:
\[ \cos x = -\frac{1}{\sqrt{2}} \]
This corresponds to:
\[ x = 2n\pi + \frac{3\pi}{4} \quad \text{or} \quad x = 2n\pi - \frac{3\pi}{4} \quad \text{where } n \in \mathbb{Z} \]

### Step 6: Combine solutions
The most general solution can be expressed as:
1. From \(\sin x = -\frac{1}{\sqrt{2}}\):
\[ x = n\pi - \frac{\pi}{4} \quad \text{where } n \in \mathbb{Z} \]
2. From \(\cos x = -\frac{1}{\sqrt{2}}\):
\[ x = 2n\pi + \frac{3\pi}{4} \quad \text{or} \quad x = 2n\pi - \frac{3\pi}{4} \quad \text{where } n \in \mathbb{Z} \]

### Final Answer
Thus, the most general solution for the equation \( 2^{\sin x} + 2^{\cos x} = 2^{1 - \frac{1}{\sqrt{2}}} \) is:
\[ x = n\pi - \frac{\pi}{4} \quad \text{or} \quad x = 2n\pi + \frac{3\pi}{4} \quad \text{or} \quad x = 2n\pi - \frac{3\pi}{4} \quad \text{where } n \in \mathbb{Z} \]