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

### Step 1: Substitute \( y = \sin x \)
We start by substituting \( y = \sin x \). The equation then becomes:
\[ 3y^2 - 7y + 2 = 0 \]

### Step 2: Solve the quadratic equation
Next, we will solve the quadratic equation using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 3 \), \( b = -7 \), and \( c = 2 \). Plugging in these values:
\[ y = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 3 \cdot 2}}{2 \cdot 3} \]
Calculating the discriminant:
\[ (-7)^2 - 4 \cdot 3 \cdot 2 = 49 - 24 = 25 \]
Now substituting back into the formula:
\[ y = \frac{7 \pm \sqrt{25}}{6} = \frac{7 \pm 5}{6} \]
This gives us two possible values for \( y \):
\[ y_1 = \frac{12}{6} = 2 \quad \text{and} \quad y_2 = \frac{2}{6} = \frac{1}{3} \]

### Step 3: Analyze the solutions for \( y \)
Since \( \sin x \) must be in the range \([-1, 1]\), we discard \( y_1 = 2 \) because it is not a valid sine value. We keep \( y_2 = \frac{1}{3} \).

### Step 4: Find the angles corresponding to \( y = \frac{1}{3} \)
Now we need to find the angles \( x \) such that \( \sin x = \frac{1}{3} \). The solutions in one full cycle (from \( 0 \) to \( 2\pi \)) are:
\[ x_1 = \sin^{-1}\left(\frac{1}{3}\right) \quad \text{and} \quad x_2 = \pi - \sin^{-1}\left(\frac{1}{3}\right) \]

### Step 5: Determine the number of solutions in the interval \( [0, 5\pi] \)
The sine function has a period of \( 2\pi \). Therefore, in each interval of \( [0, 2\pi] \), we have 2 solutions.

Now, we need to find how many complete cycles of \( 2\pi \) fit into \( [0, 5\pi] \):
- The number of complete cycles in \( [0, 5\pi] \) is \( \frac{5\pi}{2\pi} = 2.5 \), which means there are 2 complete cycles and an additional half cycle.

In each complete cycle, we have 2 solutions:
- For the first cycle \( [0, 2\pi] \): 2 solutions
- For the second cycle \( [2\pi, 4\pi] \): 2 solutions
- For the half cycle \( [4\pi, 5\pi] \): 1 solution (only \( x_1 \) is valid in this range)

### Step 6: Total number of solutions
Adding these together:
\[ 2 + 2 + 1 = 5 \]

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