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

Options

A. 0
B. 5
C. 4
D. 10

Answer

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

### Step 1: Rewrite the equation
The given equation is a quadratic in terms of \(\sin x\):
\[ 3\sin^2 x - 7\sin x + 2 = 0 \]

### Step 2: Factor the quadratic equation
We can factor this quadratic equation. We look for two numbers that multiply to \(3 \cdot 2 = 6\) and add up to \(-7\). The numbers \(-6\) and \(-1\) satisfy this condition. Thus, we can rewrite the equation as:
\[ 3\sin^2 x - 6\sin x - \sin x + 2 = 0 \]
Grouping the terms gives:
\[ 3\sin x(\sin x - 2) - 1(\sin x - 2) = 0 \]
Factoring out \((\sin x - 2)\):
\[ (3\sin x - 1)(\sin x - 2) = 0 \]

### Step 3: Solve for \(\sin x\)
Setting each factor to zero gives us:
1. \(3\sin x - 1 = 0 \implies \sin x = \frac{1}{3}\)
2. \(\sin x - 2 = 0 \implies \sin x = 2\) (not possible since the sine function has a maximum value of 1)

Thus, we only consider \(\sin x = \frac{1}{3}\).

### Step 4: Find the general solutions for \(\sin x = \frac{1}{3}\)
The general solutions for \(\sin x = k\) are given by:
\[ x = \sin^{-1}(k) + 2n\pi \quad \text{and} \quad x = \pi - \sin^{-1}(k) + 2n\pi \]
where \(n\) is any integer.

For \(\sin x = \frac{1}{3}\), we have:
\[ x_1 = \sin^{-1}\left(\frac{1}{3}\right) + 2n\pi \]
\[ x_2 = \pi - \sin^{-1}\left(\frac{1}{3}\right) + 2n\pi \]

### Step 5: Determine the values of \(x\) in the interval \([0, 3\pi]\)
Now we need to find all solutions in the interval \([0, 3\pi]\).

1. For \(n = 0\):
- \(x_1 = \sin^{-1}\left(\frac{1}{3}\right)\)
- \(x_2 = \pi - \sin^{-1}\left(\frac{1}{3}\right)\)

2. For \(n = 1\):
- \(x_1 = \sin^{-1}\left(\frac{1}{3}\right) + 2\pi\)
- \(x_2 = \pi - \sin^{-1}\left(\frac{1}{3}\right) + 2\pi\)

3. For \(n = 2\):
- \(x_1 = \sin^{-1}\left(\frac{1}{3}\right) + 4\pi\) (not in the interval)
- \(x_2 = \pi - \sin^{-1}\left(\frac{1}{3}\right) + 4\pi\) (not in the interval)

### Step 6: Count the valid solutions
Now we count the valid solutions:
- From \(n = 0\): 2 solutions
- From \(n = 1\): 2 solutions

Thus, the total number of solutions in the interval \([0, 3\pi]\) is \(2 + 2 = 4\).

### Final Answer
The number of values of \(x\) in the interval \([0, 3\pi]\) satisfying the equation is **4**.