The number of values of x in the interval `[0,5pi]` satisfying the equation `3sin^2x-7sinx+2=0` is

Options

A. 0
B. 2
C. 6
D. 4

Answer

To solve the equation \(3\sin^2 x - 7\sin x + 2 = 0\) for the number of values of \(x\) in the interval \([0, 5\pi]\), 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 the quadratic equation. We need to find two numbers that multiply to \(3 \cdot 2 = 6\) and add to \(-7\). The numbers are \(-6\) and \(-1\):
\[ 3\sin^2 x - 6\sin x - \sin x + 2 = 0 \]
Now, grouping the terms:
\[ 3\sin x(\sin x - 2) - 1(\sin x - 2) = 0 \]
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. \(3\sin x - 1 = 0 \Rightarrow \sin x = \frac{1}{3}\)
2. \(\sin x - 2 = 0 \Rightarrow \sin x = 2\)

### Step 4: Analyze the solutions
The sine function has a range of \([-1, 1]\). Thus, \(\sin x = 2\) is not a valid solution. 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:
\[ x = \arcsin(k) + 2n\pi \quad \text{and} \quad x = \pi - \arcsin(k) + 2n\pi \quad (n \in \mathbb{Z}) \]
For \(k = \frac{1}{3}\):
\[ x = \arcsin\left(\frac{1}{3}\right) + 2n\pi \quad \text{and} \quad x = \pi - \arcsin\left(\frac{1}{3}\right) + 2n\pi \]

### Step 6: Find the specific solutions in the interval \([0, 5\pi]\)
Let \(A = \arcsin\left(\frac{1}{3}\right)\):
1. First solution: \(x = A + 2n\pi\)
2. Second solution: \(x = \pi - A + 2n\pi\)

Now we find the values of \(n\) such that \(x\) is in the interval \([0, 5\pi]\).

- For \(n = 0\):
- \(x_1 = A\)
- \(x_2 = \pi - A\)

- For \(n = 1\):
- \(x_3 = A + 2\pi\)
- \(x_4 = \pi - A + 2\pi\)

- For \(n = 2\):
- \(x_5 = A + 4\pi\)
- \(x_6 = \pi - A + 4\pi\)

Now we check the values:
- \(x_1 = A\) (valid)
- \(x_2 = \pi - A\) (valid)
- \(x_3 = A + 2\pi\) (valid)
- \(x_4 = \pi - A + 2\pi\) (valid)
- \(x_5 = A + 4\pi\) (valid)
- \(x_6 = \pi - A + 4\pi\) (valid)

### Conclusion
Thus, there are a total of 6 solutions for \(x\) in the interval \([0, 5\pi]\).

**Final Answer: 6**