For `x in (-pi, pi)` find the value of `x` for which the given equation `(sqrt 3 sin x + cos x)^(sqrt(sqrt3 sin 2 x-cos 2 x+2))=4` is satisfied.

Answer

To solve the equation \((\sqrt{3} \sin x + \cos x)^{\sqrt{\sqrt{3} \sin 2x - \cos 2x + 2}} = 4\) for \(x\) in the interval \((- \pi, \pi)\), we can follow these steps:

### Step 1: Rewrite the equation
We can rewrite the equation as:
\[ 4 = (\sqrt{3} \sin x + \cos x)^{\sqrt{\sqrt{3} \sin 2x - \cos 2x + 2}} \]

### Step 2: Simplify the equation
Since \(4\) can be expressed as \(2^2\), we can rewrite the equation:
\[ (\sqrt{3} \sin x + \cos x)^{\sqrt{\sqrt{3} \sin 2x - \cos 2x + 2}} = 2^2 \]

### Step 3: Set up the base and exponent
To solve this, we can equate the base and the exponent:
1. \(\sqrt{3} \sin x + \cos x = 2\)
2. \(\sqrt{\sqrt{3} \sin 2x - \cos 2x + 2} = 2\)

### Step 4: Solve the first equation
From \(\sqrt{3} \sin x + \cos x = 2\):
- The maximum value of \(\sqrt{3} \sin x + \cos x\) occurs when \(\sin x\) and \(\cos x\) are at their maximum values. The maximum value of \(\sqrt{3} \sin x + \cos x\) is \(\sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{4} = 2\).
- Therefore, \(\sqrt{3} \sin x + \cos x = 2\) occurs when \(\sin x = 1\) and \(\cos x = 0\), which gives \(x = \frac{\pi}{2}\).

### Step 5: Solve the second equation
Now, we solve \(\sqrt{\sqrt{3} \sin 2x - \cos 2x + 2} = 2\):
- Squaring both sides gives:
\[ \sqrt{3} \sin 2x - \cos 2x + 2 = 4 \]
- Simplifying gives:
\[ \sqrt{3} \sin 2x - \cos 2x = 2 \]

### Step 6: Analyze the second equation
The maximum value of \(\sqrt{3} \sin 2x - \cos 2x\) is also determined by the maximum values of \(\sin\) and \(\cos\). The maximum value of \(\sqrt{3} \sin 2x - \cos 2x\) can be calculated similarly:
- The maximum value is \(\sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{4} = 2\).
- This occurs when \(\sin 2x = 1\) and \(\cos 2x = 0\), which gives \(2x = \frac{\pi}{2}\) or \(2x = \frac{5\pi}{2}\). Thus, \(x = \frac{\pi}{4}\) or \(x = \frac{5\pi}{4}\).

### Step 7: Check for valid solutions
1. \(x = \frac{\pi}{2}\) is valid in the interval \((- \pi, \pi)\).
2. \(x = \frac{\pi}{4}\) is also valid.
3. \(x = \frac{5\pi}{4}\) is not valid since it is outside the interval.

### Final Answer
The valid solutions for \(x\) in the interval \((- \pi, \pi)\) are:
\[ x = \frac{\pi}{2}, \frac{\pi}{4} \]