The number of solutions of the equation `cos6x+tan^2x+cos6xtan^2x=1` in the interval `[0,2pi]` is 4 (b) 5 (c) 6 (d) 7

Options

A. 4
B. 5
C. 6
D. 7

Explanation

To find the number of solutions of the equation \( \cos(6x) + \tan^2(x) + \cos(6x) \tan^2(x) = 1 \) in the interval \([0, 2\pi]\), we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ \cos(6x) + \tan^2(x) + \cos(6x) \tan^2(x) = 1 \]
We can factor out \(\cos(6x)\) from the first and third terms:
\[ \cos(6x)(1 + \tan^2(x)) + \tan^2(x) = 1 \]

### Step 2: Use the identity for \(1 + \tan^2(x)\)
Recall the identity:
\[ 1 + \tan^2(x) = \sec^2(x) \]
Substituting this into our equation gives:
\[ \cos(6x) \sec^2(x) + \tan^2(x) = 1 \]
This can be rewritten as:
\[ \cos(6x) \cdot \frac{1}{\cos^2(x)} + \tan^2(x) = 1 \]

### Step 3: Multiply through by \(\cos^2(x)\)
To eliminate the fraction, we multiply the entire equation by \(\cos^2(x)\):
\[ \cos(6x) + \tan^2(x) \cos^2(x) = \cos^2(x) \]
Using the identity \(\tan^2(x) = \frac{\sin^2(x)}{\cos^2(x)}\), we substitute:
\[ \cos(6x) + \sin^2(x) = \cos^2(x) \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ \cos(6x) = \cos^2(x) - \sin^2(x) \]
Using the identity \(\cos^2(x) - \sin^2(x) = \cos(2x)\), we have:
\[ \cos(6x) = \cos(2x) \]

### Step 5: Solve the equation \(\cos(6x) = \cos(2x)\)
The general solution for \(\cos A = \cos B\) is:
\[ A = 2n\pi \pm B \]
Applying this to our equation:
\[ 6x = 2n\pi \pm 2x \]
This gives us two cases to solve.

### Case 1: \(6x = 2n\pi + 2x\)
Rearranging gives:
\[ 4x = 2n\pi \quad \Rightarrow \quad x = \frac{n\pi}{2} \]

### Case 2: \(6x = 2n\pi - 2x\)
Rearranging gives:
\[ 8x = 2n\pi \quad \Rightarrow \quad x = \frac{n\pi}{4} \]

### Step 6: Find solutions in the interval \([0, 2\pi]\)
Now we find the values of \(n\) for both cases:

**For \(x = \frac{n\pi}{2}\):**
- \(n = 0 \Rightarrow x = 0\)
- \(n = 1 \Rightarrow x = \frac{\pi}{2}\)
- \(n = 2 \Rightarrow x = \pi\)
- \(n = 3 \Rightarrow x = \frac{3\pi}{2}\)
- \(n = 4 \Rightarrow x = 2\pi\)

This gives us 5 solutions: \(0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\).

**For \(x = \frac{n\pi}{4}\):**
- \(n = 0 \Rightarrow x = 0\)
- \(n = 1 \Rightarrow x = \frac{\pi}{4}\)
- \(n = 2 \Rightarrow x = \frac{\pi}{2}\)
- \(n = 3 \Rightarrow x = \frac{3\pi}{4}\)
- \(n = 4 \Rightarrow x = \pi\)
- \(n = 5 \Rightarrow x = \frac{5\pi}{4}\)
- \(n = 6 \Rightarrow x = \frac{3\pi}{2}\)
- \(n = 7 \Rightarrow x = \frac{7\pi}{4}\)
- \(n = 8 \Rightarrow x = 2\pi\)

This gives us 9 solutions: \(0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi, \frac{5\pi}{4}, \frac{3\pi}{2}, \frac{7\pi}{4}, 2\pi\).

### Step 7: Combine and count unique solutions
Combining both cases, we have:
- From \(x = \frac{n\pi}{2}\): \(0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\)
- From \(x = \frac{n\pi}{4}\): \(0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi, \frac{5\pi}{4}, \frac{3\pi}{2}, \frac{7\pi}{4}, 2\pi\)

The unique solutions are:
\[ 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi, \frac{5\pi}{4}, \frac{3\pi}{2}, \frac{7\pi}{4}, 2\pi \]
This gives us a total of 9 unique solutions.

### Conclusion
The number of solutions of the equation \( \cos(6x) + \tan^2(x) + \cos(6x) \tan^2(x) = 1 \) in the interval \([0, 2\pi]\) is **9**.