Number of solutions of equation `2"sin" x/2 cos^(2) x-2 "sin" x/2 sin^(2) x=cos^(2) x-sin^(2) x` for `x in [0, 4pi]` is

Options

A. 6
B. 8
C. 10
D. 12

Explanation

To solve the equation \( 2 \sin \frac{x}{2} \cos^2 x - 2 \sin \frac{x}{2} \sin^2 x = \cos^2 x - \sin^2 x \) for \( x \) in the interval \( [0, 4\pi] \), we can follow these steps:

### Step 1: Factor out common terms
We can factor out \( 2 \sin \frac{x}{2} \) from the left-hand side:
\[ 2 \sin \frac{x}{2} (\cos^2 x - \sin^2 x) = \cos^2 x - \sin^2 x \]

### Step 2: Rearrange the equation
Now we can rearrange the equation:
\[ 2 \sin \frac{x}{2} (\cos^2 x - \sin^2 x) - (\cos^2 x - \sin^2 x) = 0 \]
This simplifies to:
\[ (\cos^2 x - \sin^2 x)(2 \sin \frac{x}{2} - 1) = 0 \]

### Step 3: Solve each factor
We have two factors to solve:
1. \( \cos^2 x - \sin^2 x = 0 \)
2. \( 2 \sin \frac{x}{2} - 1 = 0 \)

#### Factor 1: \( \cos^2 x - \sin^2 x = 0 \)
This can be rewritten using the identity \( \cos 2x = 0 \):
\[ \cos 2x = 0 \]
The solutions for \( 2x \) are:
\[ 2x = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
Thus,
\[ x = \frac{\pi}{4} + \frac{n\pi}{2} \]

Now, we need to find the values of \( x \) in the interval \( [0, 4\pi] \):
- For \( n = 0 \): \( x = \frac{\pi}{4} \)
- For \( n = 1 \): \( x = \frac{3\pi}{4} \)
- For \( n = 2 \): \( x = \frac{5\pi}{4} \)
- For \( n = 3 \): \( x = \frac{7\pi}{4} \)
- For \( n = 4 \): \( x = \frac{9\pi}{4} \)
- For \( n = 5 \): \( x = \frac{11\pi}{4} \)
- For \( n = 6 \): \( x = \frac{13\pi}{4} \)
- For \( n = 7 \): \( x = \frac{15\pi}{4} \)

This gives us 8 solutions from this factor.

#### Factor 2: \( 2 \sin \frac{x}{2} - 1 = 0 \)
Solving this gives:
\[ \sin \frac{x}{2} = \frac{1}{2} \]
The solutions for \( \frac{x}{2} \) are:
\[ \frac{x}{2} = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad \frac{x}{2} = \frac{5\pi}{6} + 2n\pi \]
Thus,
\[ x = \frac{\pi}{3} + 4n\pi \quad \text{or} \quad x = \frac{5\pi}{3} + 4n\pi \]

Finding values in the interval \( [0, 4\pi] \):
- For \( n = 0 \): \( x = \frac{\pi}{3} \) and \( x = \frac{5\pi}{3} \)
- For \( n = 1 \): \( x = \frac{13\pi}{3} \) (which is \( \frac{13\pi}{3} > 4\pi \))

This gives us 2 solutions from this factor.

### Step 4: Total solutions
Adding the solutions from both factors:
- From \( \cos^2 x - \sin^2 x = 0 \): 8 solutions
- From \( 2 \sin \frac{x}{2} - 1 = 0 \): 2 solutions

Thus, the total number of solutions is:
\[ 8 + 2 = 10 \]

### Final Answer
The number of solutions of the equation in the interval \( [0, 4\pi] \) is \( \boxed{10} \).