The number of solutions of the equation `cos^(2)((pi)/(3)cos x - (8pi)/(3))=1` in the interval `[0,10pi]` is

Options

A. 1
B. 3
C. 5
D. 7

Explanation

To solve the equation \( \cos^2\left(\frac{\pi}{3} \cos x - \frac{8\pi}{3}\right) = 1 \) in the interval \([0, 10\pi]\), we will follow these steps:

### Step 1: Simplify the Equation
The equation \( \cos^2\left(\frac{\pi}{3} \cos x - \frac{8\pi}{3}\right) = 1 \) implies that:
\[ \cos\left(\frac{\pi}{3} \cos x - \frac{8\pi}{3}\right) = \pm 1 \]
This means that the angle \( \frac{\pi}{3} \cos x - \frac{8\pi}{3} \) must be equal to \( 2n\pi \) or \( (2n + 1)\pi \) for some integer \( n \).

### Step 2: Set Up the Equations
From the above, we have two cases:
1. \( \frac{\pi}{3} \cos x - \frac{8\pi}{3} = 2n\pi \)
2. \( \frac{\pi}{3} \cos x - \frac{8\pi}{3} = (2n + 1)\pi \)

### Step 3: Solve the First Case
For the first case:
\[ \frac{\pi}{3} \cos x - \frac{8\pi}{3} = 2n\pi \]
Rearranging gives:
\[ \frac{\pi}{3} \cos x = 2n\pi + \frac{8\pi}{3} \]
\[ \cos x = 6n + 8 \]

### Step 4: Determine Valid Values for \( n \)
Since \( \cos x \) must be in the range \([-1, 1]\):
\[ -1 \leq 6n + 8 \leq 1 \]
Solving the inequalities:
1. \( 6n + 8 \geq -1 \) gives \( 6n \geq -9 \) or \( n \geq -\frac{3}{2} \)
2. \( 6n + 8 \leq 1 \) gives \( 6n \leq -7 \) or \( n \leq -\frac{7}{6} \)

The integer values of \( n \) that satisfy both inequalities are \( n = -1 \).

### Step 5: Solve the Second Case
For the second case:
\[ \frac{\pi}{3} \cos x - \frac{8\pi}{3} = (2n + 1)\pi \]
Rearranging gives:
\[ \frac{\pi}{3} \cos x = (2n + 1)\pi + \frac{8\pi}{3} \]
\[ \cos x = 3(2n + 1) + 8 \]
\[ \cos x = 6n + 3 + 8 = 6n + 11 \]

### Step 6: Determine Valid Values for \( n \)
Again, since \( \cos x \) must be in the range \([-1, 1]\):
\[ -1 \leq 6n + 11 \leq 1 \]
Solving the inequalities:
1. \( 6n + 11 \geq -1 \) gives \( 6n \geq -12 \) or \( n \geq -2 \)
2. \( 6n + 11 \leq 1 \) gives \( 6n \leq -10 \) or \( n \leq -\frac{5}{3} \)

The integer values of \( n \) that satisfy both inequalities are \( n = -2 \).

### Step 7: Find Solutions for Each Case
1. For \( n = -1 \):
\[ \cos x = 6(-1) + 8 = 2 \quad \text{(not valid)} \]

2. For \( n = -2 \):
\[ \cos x = 6(-2) + 11 = -1 \]
This gives \( x = \pi + 2k\pi \) for integers \( k \).

### Step 8: Count the Solutions in the Interval \([0, 10\pi]\)
The solutions \( x = \pi + 2k\pi \) can be expressed as:
\[ x = (2k + 1)\pi \]
For \( k = 0, 1, 2, 3, 4 \):
- \( k = 0 \): \( x = \pi \)
- \( k = 1 \): \( x = 3\pi \)
- \( k = 2 \): \( x = 5\pi \)
- \( k = 3 \): \( x = 7\pi \)
- \( k = 4 \): \( x = 9\pi \)

Thus, there are **5 solutions** in the interval \([0, 10\pi]\).

### Final Answer
The number of solutions of the equation in the interval \([0, 10\pi]\) is **5**.