The number of solutions of the equation
`3"sin"^(2) x - 7"sin" x +2 = 0`
in the interval `[0, 5 pi]`, is

Options

A. 0
B. 5
C. 6
D. 10

Explanation

To solve the equation \( 3\sin^2 x - 7\sin x + 2 = 0 \) and find the number of solutions in the interval \([0, 5\pi]\), we can follow these steps:

### Step 1: Substitute \( y = \sin x \)
We rewrite the equation in terms of \( y \):
\[ 3y^2 - 7y + 2 = 0 \]

### Step 2: Factor the quadratic equation
To factor the quadratic equation, we can rewrite it as:
\[ 3y^2 - 6y - y + 2 = 0 \]
Grouping the terms gives us:
\[ (3y^2 - 6y) + (-y + 2) = 0 \]
Factoring out common terms:
\[ 3y(y - 2) - 1(y - 2) = 0 \]
Now we can factor out \( (y - 2) \):
\[ (y - 2)(3y - 1) = 0 \]

### Step 3: Solve for \( y \)
Setting each factor to zero gives us:
1. \( y - 2 = 0 \) → \( y = 2 \)
2. \( 3y - 1 = 0 \) → \( y = \frac{1}{3} \)

### Step 4: Analyze the solutions
Since \( y = \sin x \) must satisfy \( -1 \leq \sin x \leq 1 \), we discard \( y = 2 \) because it is outside the range of the sine function. Thus, we only consider:
\[ \sin x = \frac{1}{3} \]

### Step 5: Find the general solutions for \( \sin x = \frac{1}{3} \)
The general solutions for \( \sin x = k \) are given by:
\[ x = \arcsin(k) + 2n\pi \quad \text{and} \quad x = \pi - \arcsin(k) + 2n\pi \]
For \( k = \frac{1}{3} \):
1. \( x = \arcsin\left(\frac{1}{3}\right) + 2n\pi \)
2. \( x = \pi - \arcsin\left(\frac{1}{3}\right) + 2n\pi \)

### Step 6: Determine the number of solutions in the interval \([0, 5\pi]\)
To find the number of solutions, we need to determine how many times these solutions fit into the interval \([0, 5\pi]\).

1. **For \( x = \arcsin\left(\frac{1}{3}\right) + 2n\pi \)**:
- The first solution is \( \arcsin\left(\frac{1}{3}\right) \).
- The next solutions are \( \arcsin\left(\frac{1}{3}\right) + 2\pi \) and \( \arcsin\left(\frac{1}{3}\right) + 4\pi \).

2. **For \( x = \pi - \arcsin\left(\frac{1}{3}\right) + 2n\pi \)**:
- The first solution is \( \pi - \arcsin\left(\frac{1}{3}\right) \).
- The next solutions are \( \pi - \arcsin\left(\frac{1}{3}\right) + 2\pi \) and \( \pi - \arcsin\left(\frac{1}{3}\right) + 4\pi \).

### Step 7: Count the solutions
Now, we count the number of valid solutions:
- For \( n = 0 \): \( \arcsin\left(\frac{1}{3}\right) \) and \( \pi - \arcsin\left(\frac{1}{3}\right) \) (2 solutions)
- For \( n = 1 \): \( \arcsin\left(\frac{1}{3}\right) + 2\pi \) and \( \pi - \arcsin\left(\frac{1}{3}\right) + 2\pi \) (2 solutions)
- For \( n = 2 \): \( \arcsin\left(\frac{1}{3}\right) + 4\pi \) and \( \pi - \arcsin\left(\frac{1}{3}\right) + 4\pi \) (2 solutions)

Thus, we have a total of \( 2 + 2 + 2 = 6 \) solutions in the interval \([0, 5\pi]\).

### Final Answer
The number of solutions of the equation \( 3\sin^2 x - 7\sin x + 2 = 0 \) in the interval \([0, 5\pi]\) is **6**.

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