The number of distinct solutions of the equation `5/4cos^(2)2x + cos^4 x + sin^4 x+cos^6x+sin^6 x =2` in the interval `[0,2pi] ` is

Explanation

To solve the equation \( \frac{5}{4} \cos^2 2x + \cos^4 x + \sin^4 x + \cos^6 x + \sin^6 x = 2 \) in the interval \( [0, 2\pi] \), we will follow these steps:

### Step 1: Simplify the Equation
We start with the equation:
\[ \frac{5}{4} \cos^2 2x + \cos^4 x + \sin^4 x + \cos^6 x + \sin^6 x = 2 \]

Notice that \( \cos^4 x + \sin^4 x \) can be rewritten using the identity:
\[ \cos^4 x + \sin^4 x = (\cos^2 x + \sin^2 x)^2 - 2\cos^2 x \sin^2 x = 1 - 2\cos^2 x \sin^2 x \]

Also, \( \cos^6 x + \sin^6 x \) can be expressed as:
\[ \cos^6 x + \sin^6 x = (\cos^2 x + \sin^2 x)(\cos^4 x + \sin^4 x - \cos^2 x \sin^2 x) = 1 \cdot (1 - 2\cos^2 x \sin^2 x - \cos^2 x \sin^2 x) = 1 - 3\cos^2 x \sin^2 x \]

Substituting these into our equation gives:
\[ \frac{5}{4} \cos^2 2x + 1 - 2\cos^2 x \sin^2 x + 1 - 3\cos^2 x \sin^2 x = 2 \]

### Step 2: Combine Like Terms
Now we combine like terms:
\[ \frac{5}{4} \cos^2 2x + 2 - 5\cos^2 x \sin^2 x = 2 \]
This simplifies to:
\[ \frac{5}{4} \cos^2 2x - 5\cos^2 x \sin^2 x = 0 \]

### Step 3: Factor the Equation
Factoring out \( 5 \):
\[ 5 \left( \frac{1}{4} \cos^2 2x - \cos^2 x \sin^2 x \right) = 0 \]
This implies:
\[ \frac{1}{4} \cos^2 2x = \cos^2 x \sin^2 x \]

### Step 4: Use Trigonometric Identities
Using the identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \) and \( \cos^2 x = \frac{1 + \cos 2x}{2} \), we can express \( \cos^2 x \sin^2 x \):
\[ \cos^2 x \sin^2 x = \left(\frac{1 + \cos 2x}{2}\right) \left(\frac{1 - \cos 2x}{2}\right) = \frac{1 - \cos^2 2x}{4} = \frac{\sin^2 2x}{4} \]

### Step 5: Substitute Back
Substituting back into our equation gives:
\[ \frac{1}{4} \cos^2 2x = \frac{\sin^2 2x}{4} \]
Multiplying through by 4:
\[ \cos^2 2x = \sin^2 2x \]

### Step 6: Solve for \( 2x \)
This leads to:
\[ \tan^2 2x = 1 \implies \tan 2x = \pm 1 \]
Thus:
\[ 2x = n\pi + \frac{\pi}{4} \quad \text{or} \quad 2x = n\pi - \frac{\pi}{4}, \quad n \in \mathbb{Z} \]

### Step 7: Solve for \( x \)
Dividing by 2:
\[ x = \frac{n\pi}{2} + \frac{\pi}{8} \quad \text{or} \quad x = \frac{n\pi}{2} - \frac{\pi}{8} \]

### Step 8: Find Distinct Solutions in \( [0, 2\pi] \)
Now we find the values of \( x \) in the interval \( [0, 2\pi] \):
1. For \( x = \frac{n\pi}{2} + \frac{\pi}{8} \):
- \( n = 0 \): \( x = \frac{\pi}{8} \)
- \( n = 1 \): \( x = \frac{5\pi}{8} \)
- \( n = 2 \): \( x = \frac{9\pi}{8} \)
- \( n = 3 \): \( x = \frac{13\pi}{8} \)

2. For \( x = \frac{n\pi}{2} - \frac{\pi}{8} \):
- \( n = 0 \): \( x = -\frac{\pi}{8} \) (not in interval)
- \( n = 1 \): \( x = \frac{3\pi}{8} \)
- \( n = 2 \): \( x = \frac{7\pi}{8} \)
- \( n = 3 \): \( x = \frac{11\pi}{8} \)
- \( n = 4 \): \( x = \frac{15\pi}{8} \)

### Step 9: Count Distinct Solutions
The distinct solutions in \( [0, 2\pi] \) are:
- \( \frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8}, \frac{9\pi}{8}, \frac{11\pi}{8}, \frac{13\pi}{8}, \frac{15\pi}{8} \)

Thus, there are a total of 8 distinct solutions.

### Final Answer
The number of distinct solutions is \( \boxed{8} \).