Number of solutions of equation `sin x.sqrt(8cos^(2)x)=1` in `[0, 2pi]` are

Explanation

To find the number of solutions of the equation \( \sin x \sqrt{8 \cos^2 x} = 1 \) in the interval \( [0, 2\pi] \), we can follow these steps:

### Step 1: Rewrite the equation
Start with the given equation:
\[ \sin x \sqrt{8 \cos^2 x} = 1 \]
We can rewrite \( \sqrt{8 \cos^2 x} \) as \( 2\sqrt{2} |\cos x| \). Therefore, the equation becomes:
\[ \sin x \cdot 2\sqrt{2} |\cos x| = 1 \]
This simplifies to:
\[ 2\sqrt{2} \sin x |\cos x| = 1 \]

### Step 2: Analyze the absolute value
Since \( |\cos x| \) depends on the quadrant of \( x \), we will consider the intervals where \( \cos x \) is positive and negative.

1. **For \( x \in [0, \frac{\pi}{2}] \)** and **\( x \in [\frac{3\pi}{2}, 2\pi] \)**, \( \cos x \) is positive.
2. **For \( x \in [\frac{\pi}{2}, \frac{3\pi}{2}] \)**, \( \cos x \) is negative.

### Step 3: Solve for each case
#### Case 1: \( x \in [0, \frac{\pi}{2}] \)
Here, \( |\cos x| = \cos x \):
\[ 2\sqrt{2} \sin x \cos x = 1 \]
Using the identity \( \sin 2x = 2 \sin x \cos x \), we rewrite the equation as:
\[ \sqrt{2} \sin 2x = 1 \]
Thus,
\[ \sin 2x = \frac{1}{\sqrt{2}} \]
The solutions for \( \sin \theta = \frac{1}{\sqrt{2}} \) are:
\[ 2x = \frac{\pi}{4} + 2k\pi \quad \text{and} \quad 2x = \frac{3\pi}{4} + 2k\pi \]
For \( k = 0 \):
\[ x = \frac{\pi}{8} \quad \text{and} \quad x = \frac{3\pi}{8} \]

#### Case 2: \( x \in [\frac{\pi}{2}, \frac{3\pi}{2}] \)
Here, \( |\cos x| = -\cos x \):
\[ 2\sqrt{2} \sin x (-\cos x) = 1 \]
This simplifies to:
\[ -2\sqrt{2} \sin x \cos x = 1 \]
Thus,
\[ \sin 2x = -\frac{1}{\sqrt{2}} \]
The solutions for \( \sin \theta = -\frac{1}{\sqrt{2}} \) are:
\[ 2x = \frac{5\pi}{4} + 2k\pi \quad \text{and} \quad 2x = \frac{7\pi}{4} + 2k\pi \]
For \( k = 0 \):
\[ x = \frac{5\pi}{8} \quad \text{and} \quad x = \frac{7\pi}{8} \]

#### Case 3: \( x \in [\frac{3\pi}{2}, 2\pi] \)
Here, \( |\cos x| = \cos x \):
\[ 2\sqrt{2} \sin x \cos x = 1 \]
This case is the same as Case 1, leading to:
\[ \sin 2x = \frac{1}{\sqrt{2}} \]
The solutions will again be:
\[ x = \frac{\pi}{8} \quad \text{and} \quad x = \frac{3\pi}{8} \]

### Step 4: Count the distinct solutions
From the three cases, we have the following solutions:
1. From Case 1: \( x = \frac{\pi}{8}, \frac{3\pi}{8} \)
2. From Case 2: \( x = \frac{5\pi}{8}, \frac{7\pi}{8} \)
3. Case 3 does not yield new solutions since it repeats Case 1.

### Final Answer
Thus, the total number of distinct solutions in the interval \( [0, 2\pi] \) is:
\[ \boxed{4} \]