The number of solutions of the equation
`|2 sin x-sqrt(3)|^(2 cos^(2) x-3 cos x+1)=1` in `[0, pi]` is

Options

A. 2
B. 3
C. 4
D. 5

Explanation

To solve the equation

\[ |2 \sin x - \sqrt{3}|^{(2 \cos^2 x - 3 \cos x + 1)} = 1 \]

in the interval \([0, \pi]\), we can break it down into two cases based on the properties of exponents.

### Step 1: Identify the conditions for the equation to hold
The equation \(a^b = 1\) holds true under two conditions:
1. \(a = 1\)
2. \(b = 0\) (and \(a \neq 0\))

### Step 2: Solve for the first condition \(a = 1\)
Set the base equal to 1:
\[ |2 \sin x - \sqrt{3}| = 1 \]

This gives us two equations to solve:
1. \(2 \sin x - \sqrt{3} = 1\)
2. \(2 \sin x - \sqrt{3} = -1\)

**For the first equation:**
\[ 2 \sin x = 1 + \sqrt{3} \]
\[ \sin x = \frac{1 + \sqrt{3}}{2} \]
Since \(\sqrt{3} \approx 1.732\), we have:
\[ \sin x \approx \frac{1 + 1.732}{2} = \frac{2.732}{2} \approx 1.366 \]
This value is not possible since \(\sin x\) cannot exceed 1.

**For the second equation:**
\[ 2 \sin x = -1 + \sqrt{3} \]
\[ \sin x = \frac{-1 + \sqrt{3}}{2} \]
Calculating this gives:
\[ \sin x \approx \frac{-1 + 1.732}{2} = \frac{0.732}{2} \approx 0.366 \]
This value is valid.

### Step 3: Find the angles corresponding to \(\sin x = 0.366\)
In the interval \([0, \pi]\), \(\sin x = 0.366\) gives us two solutions:
1. \(x_1 = \arcsin(0.366)\)
2. \(x_2 = \pi - \arcsin(0.366)\)

### Step 4: Solve for the second condition \(b = 0\)
Set the exponent to zero:
\[ 2 \cos^2 x - 3 \cos x + 1 = 0 \]
This is a quadratic equation in terms of \(\cos x\). We can factor it:
\[ (2 \cos x - 1)(\cos x - 1) = 0 \]
This gives us:
1. \(2 \cos x - 1 = 0 \Rightarrow \cos x = \frac{1}{2}\)
2. \(\cos x - 1 = 0 \Rightarrow \cos x = 1\)

**Finding the angles:**
- For \(\cos x = \frac{1}{2}\):
\[ x = \frac{\pi}{3} \]
- For \(\cos x = 1\):
\[ x = 0 \]

### Step 5: Verify the solutions
Now we have the following potential solutions:
1. \(x = 0\)
2. \(x = \frac{\pi}{3}\)
3. \(x = \arcsin(0.366)\)
4. \(x = \pi - \arcsin(0.366)\)

### Step 6: Check if any solutions are extraneous
We need to check if any of these solutions make the original equation undefined or invalid.

- For \(x = 0\):
\[ |2 \sin(0) - \sqrt{3}|^{(2 \cos^2(0) - 3 \cos(0) + 1)} = |-\sqrt{3}|^{(2 - 3 + 1)} = \sqrt{3}^0 = 1 \]
- For \(x = \frac{\pi}{3}\):
\[ |2 \sin(\frac{\pi}{3}) - \sqrt{3}|^{(2 \cos^2(\frac{\pi}{3}) - 3 \cos(\frac{\pi}{3}) + 1)} = |2 \cdot \frac{\sqrt{3}}{2} - \sqrt{3}|^{(2 \cdot \frac{1}{4} - 3 \cdot \frac{1}{2} + 1)} = |0|^{0} \text{ (undefined)} \]
- For \(x = \arcsin(0.366)\) and \(x = \pi - \arcsin(0.366)\), we need to check if they yield valid outputs, but since they are derived from valid sine values, they should also hold.

### Conclusion
Thus, the valid solutions in the interval \([0, \pi]\) are:
1. \(x = 0\)
2. \(x = \arcsin(0.366)\)
3. \(x = \pi - \arcsin(0.366)\)

This gives us a total of **three solutions**.

### Final Answer
The number of solutions of the equation in \([0, \pi]\) is **3**.

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