Number of solution(s) satisfying the equation `1/(sinx)-1/(sin2x)=2/(sin4x)` in `[0,4pi]` equals 0 (b) 2 (c) 4 (d) 6

Options

A. 0
B. 2
C. 4
D. 6

Explanation

To solve the equation \( \frac{1}{\sin x} - \frac{1}{\sin 2x} = \frac{2}{\sin 4x} \) in the interval \([0, 4\pi]\), we will follow these steps:

### Step 1: Rewrite the equation
Start by rewriting the equation:
\[ \frac{1}{\sin x} - \frac{1}{\sin 2x} = \frac{2}{\sin 4x} \]

### Step 2: Find a common denominator
To combine the left-hand side, we need a common denominator:
\[ \frac{\sin 2x - \sin x}{\sin x \sin 2x} = \frac{2}{\sin 4x} \]

### Step 3: Use the identity for \(\sin 4x\)
Recall the double angle identity:
\[ \sin 4x = 2 \sin 2x \cos 2x \]
Substituting this into the equation gives:
\[ \frac{\sin 2x - \sin x}{\sin x \sin 2x} = \frac{2}{2 \sin 2x \cos 2x} \]
This simplifies to:
\[ \frac{\sin 2x - \sin x}{\sin x \sin 2x} = \frac{1}{\sin 2x \cos 2x} \]

### Step 4: Cross-multiply
Cross-multiplying yields:
\[ (\sin 2x - \sin x) \cos 2x = \sin x \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ \sin 2x \cos 2x - \sin x \cos 2x - \sin x = 0 \]
This can be factored:
\[ \sin 2x \cos 2x - \sin x (\cos 2x + 1) = 0 \]

### Step 6: Solve the factors
This gives us two cases to solve:
1. \( \sin 2x \cos 2x = 0 \)
2. \( \sin x (\cos 2x + 1) = 0 \)

### Step 7: Solve \( \sin 2x \cos 2x = 0 \)
This implies:
\[ \sin 2x = 0 \quad \text{or} \quad \cos 2x = 0 \]
- For \( \sin 2x = 0 \):
\[ 2x = n\pi \implies x = \frac{n\pi}{2} \]
In the interval \([0, 4\pi]\), \( n = 0, 1, 2, 3, 4, 5, 6, 7, 8\) gives solutions: \(0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi, \frac{5\pi}{2}, 3\pi, \frac{7\pi}{2}, 4\pi\).

- For \( \cos 2x = 0 \):
\[ 2x = \frac{\pi}{2} + n\pi \implies x = \frac{\pi}{4} + \frac{n\pi}{2} \]
In the interval \([0, 4\pi]\), \( n = 0, 1, 2, 3, 4, 5, 6, 7\) gives solutions: \(\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \frac{9\pi}{4}, \frac{11\pi}{4}, \frac{13\pi}{4}, \frac{15\pi}{4}\).

### Step 8: Solve \( \sin x (\cos 2x + 1) = 0 \)
This implies:
1. \( \sin x = 0 \) gives \( x = n\pi \) for \( n = 0, 1, 2, 3, 4 \) which gives \( 0, \pi, 2\pi, 3\pi, 4\pi \).
2. \( \cos 2x + 1 = 0 \) gives no additional solutions since \( \cos 2x = -1 \) leads to \( 2x = (2n+1)\pi \) which does not yield new solutions in the given interval.

### Step 9: Count unique solutions
Now we count the unique solutions from all cases:
- From \( \sin 2x = 0 \): \( 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi, \frac{5\pi}{2}, 3\pi, \frac{7\pi}{2}, 4\pi \) (9 solutions).
- From \( \cos 2x = 0 \): \( \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \frac{9\pi}{4}, \frac{11\pi}{4}, \frac{13\pi}{4}, \frac{15\pi}{4} \) (8 solutions).
- From \( \sin x = 0 \): \( 0, \pi, 2\pi, 3\pi, 4\pi \) (5 solutions).

### Final Count
Upon combining and eliminating duplicates, we find that there are a total of **4 unique solutions** in the interval \([0, 4\pi]\).

### Conclusion
Thus, the number of solutions satisfying the equation is \( \boxed{4} \).