Solve the equation for ` 0 le x le 2pi.`
`sin x cos x =0`

Answer

To solve the equation \( \sin x \cos x = 0 \) for \( 0 \leq x \leq 2\pi \), we can follow these steps:

### Step 1: Set the equation to zero
The equation is given as:
\[ \sin x \cos x = 0 \]
This implies that either \( \sin x = 0 \) or \( \cos x = 0 \).

### Step 2: Solve \( \sin x = 0 \)
The sine function is equal to zero at:
\[ x = n\pi \]
where \( n \) is an integer. Within the range \( 0 \leq x \leq 2\pi \), the values of \( n \) that satisfy this condition are:
- \( n = 0 \): \( x = 0 \)
- \( n = 1 \): \( x = \pi \)
- \( n = 2 \): \( x = 2\pi \)

Thus, the solutions from \( \sin x = 0 \) are:
\[ x = 0, \pi, 2\pi \]

### Step 3: Solve \( \cos x = 0 \)
The cosine function is equal to zero at:
\[ x = \frac{\pi}{2} + n\pi \]
where \( n \) is an integer. Within the range \( 0 \leq x \leq 2\pi \), the values of \( n \) that satisfy this condition are:
- \( n = 0 \): \( x = \frac{\pi}{2} \)
- \( n = 1 \): \( x = \frac{3\pi}{2} \)

Thus, the solutions from \( \cos x = 0 \) are:
\[ x = \frac{\pi}{2}, \frac{3\pi}{2} \]

### Step 4: Combine the solutions
Now, we combine all the solutions obtained from both conditions:
\[ x = 0, \pi, 2\pi, \frac{\pi}{2}, \frac{3\pi}{2} \]

### Final Answer
The complete set of solutions for the equation \( \sin x \cos x = 0 \) in the interval \( 0 \leq x \leq 2\pi \) is:
\[ x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi \]