The number of solution of `sin^(4)x-cos^(2) x sin x+2 sin^(2)x+sin x=0` in `0 le x le 3 pi` is

Options

A. 3
B. 4
C. 5
D. 6

Explanation

To solve the equation \( \sin^4 x - \cos^2 x \sin x + 2 \sin^2 x + \sin x = 0 \) in the interval \( 0 \leq x \leq 3\pi \), we will follow these steps:

### Step 1: Rewrite the equation
The given equation is:
\[ \sin^4 x - \cos^2 x \sin x + 2 \sin^2 x + \sin x = 0 \]

### Step 2: Factor out common terms
We can factor out \( \sin x \) from the terms that contain it. First, let's rewrite \( \cos^2 x \) using the identity \( \cos^2 x = 1 - \sin^2 x \):
\[ \sin^4 x + 2 \sin^2 x + \sin x - (1 - \sin^2 x) \sin x = 0 \]
This simplifies to:
\[ \sin^4 x + 2 \sin^2 x + \sin x - \sin x + \sin^3 x = 0 \]
Now we can factor out \( \sin x \):
\[ \sin x (\sin^3 x + \sin^2 x + 2) = 0 \]

### Step 3: Set each factor to zero
Now we have two factors:
1. \( \sin x = 0 \)
2. \( \sin^3 x + \sin^2 x + 2 = 0 \)

### Step 4: Solve \( \sin x = 0 \)
The solutions for \( \sin x = 0 \) in the interval \( 0 \leq x \leq 3\pi \) are:
\[ x = 0, \pi, 2\pi, 3\pi \]
This gives us 4 solutions.

### Step 5: Analyze \( \sin^3 x + \sin^2 x + 2 = 0 \)
Next, we need to check if the second equation \( \sin^3 x + \sin^2 x + 2 = 0 \) has any solutions.

Notice that:
- The term \( \sin^3 x + \sin^2 x \) is always non-negative since \( \sin^2 x \) is between 0 and 1.
- Thus, \( \sin^3 x + \sin^2 x + 2 \) is always greater than or equal to 2.

Since \( \sin^3 x + \sin^2 x + 2 \) cannot equal 0, there are no solutions from this factor.

### Conclusion
The only solutions come from \( \sin x = 0 \), which gives us the solutions:
\[ x = 0, \pi, 2\pi, 3\pi \]
Thus, the total number of solutions in the interval \( 0 \leq x \leq 3\pi \) is **4**.