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 find the number of solutions of 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 will 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: Set the equation to zero
Rearranging gives us:

\[ 2 \sin \frac{x}{2} (\cos^2 x - \sin^2 x) - (\cos^2 x - \sin^2 x) = 0 \]

This can be simplified to:

\[ (\cos^2 x - \sin^2 x)(2 \sin \frac{x}{2} - 1) = 0 \]

### Step 3: Solve the factors separately
We have two factors to consider:

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:

\[ \cos 2x = 0 \]

The solutions to \(\cos 2x = 0\) are given by:

\[ 2x = \frac{\pi}{2} + n\pi \quad \Rightarrow \quad x = \frac{\pi}{4} + \frac{n\pi}{2} \]

For \(n = 0, 1, 2, 3, 4, 5, 6, 7\), we 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 \(\sin \frac{x}{2} = \frac{1}{2}\) are:

\[ \frac{x}{2} = \frac{\pi}{6} + 2n\pi \quad \Rightarrow \quad x = \frac{\pi}{3} + 4n\pi \]
\[ \frac{x}{2} = \frac{5\pi}{6} + 2n\pi \quad \Rightarrow \quad x = \frac{5\pi}{3} + 4n\pi \]

For \(n = 0\), we find:

- \(x = \frac{\pi}{3}\)
- \(x = \frac{5\pi}{3}\)

Thus, we have 2 additional solutions from this factor.

### Step 4: Count total solutions
Adding the solutions from both factors, we have:

- From \(\cos^2 x - \sin^2 x = 0\): 8 solutions
- From \(2 \sin \frac{x}{2} - 1 = 0\): 2 solutions

Total number of solutions:

\[ 8 + 2 = 10 \]

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