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

Explanation

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

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

### Step 2: Factor Out Common Terms
Notice that we can factor out \( \sin x \) from the equation:
\[ \sin x (\sin^3 x - \cos^2 x + 2 \sin x + 1) = 0 \]

### Step 3: Set Each Factor to Zero
This gives us two cases to consider:
1. \( \sin x = 0 \)
2. \( \sin^3 x - \cos^2 x + 2 \sin x + 1 = 0 \)

### Step 4: Solve the First Case
For \( \sin x = 0 \):
\[ x = n\pi \quad (n \in \mathbb{Z}) \]
In the interval \( 0 \leq x \leq 3\pi \), the solutions are:
- \( x = 0 \)
- \( x = \pi \)
- \( x = 2\pi \)
- \( x = 3\pi \)

Thus, we have **4 solutions** from this case.

### Step 5: Solve the Second Case
Now, consider the second case:
\[ \sin^3 x - \cos^2 x + 2 \sin x + 1 = 0 \]
Using the identity \( \cos^2 x = 1 - \sin^2 x \), we can rewrite the equation:
\[ \sin^3 x - (1 - \sin^2 x) + 2 \sin x + 1 = 0 \]
This simplifies to:
\[ \sin^3 x + \sin^2 x + 2 \sin x + 1 = 0 \]

### Step 6: Analyze the Polynomial
Let \( y = \sin x \). The equation becomes:
\[ y^3 + y^2 + 2y + 1 = 0 \]
We can check for rational roots using the Rational Root Theorem. Testing \( y = -1 \):
\[ (-1)^3 + (-1)^2 + 2(-1) + 1 = -1 + 1 - 2 + 1 = -1 \quad \text{(not a root)} \]
Testing \( y = -1 \) again shows it is not a root. We can also check for other roots or use numerical methods, but this polynomial does not yield any real roots in the range of \( y = \sin x \) (which is between -1 and 1).

### Step 7: Conclusion
Since the second case does not yield any additional solutions, we conclude that the only solutions come from the first case.

Thus, the total number of solutions in the interval \( 0 \leq x \leq 3\pi \) is:
\[ \boxed{4} \]