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

Options

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

Explanation

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

### Step 1: Substitute \(t = \sin x\)
We start by letting \(t = \sin x\). This transforms our equation into a standard quadratic form:
\[ 3t^2 - 7t + 2 = 0 \]

### Step 2: Factor the quadratic equation
Next, we need to factor the quadratic equation. We look for two numbers that multiply to \(3 \times 2 = 6\) and add to \(-7\). The numbers \(-6\) and \(-1\) work:
\[ 3t^2 - 6t - t + 2 = 0 \]
Now we can group the terms:
\[ 3t(t - 2) - 1(t - 2) = 0 \]
Factoring out the common term \((t - 2)\):
\[ (t - 2)(3t - 1) = 0 \]

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

### Step 4: Analyze the solutions
Since \(t = \sin x\), we need to check the validity of these solutions:
1. \(t = 2\) is not possible because the sine function only takes values in the range \([-1, 1]\).
2. \(t = \frac{1}{3}\) is valid since it lies within the range of the sine function.

### Step 5: Find the values of \(x\) for \(t = \frac{1}{3}\)
The equation \(\sin x = \frac{1}{3}\) will have solutions in the intervals:
- First quadrant: \(x = \arcsin\left(\frac{1}{3}\right)\)
- Second quadrant: \(x = \pi - \arcsin\left(\frac{1}{3}\right)\)

### Step 6: Determine the number of solutions in \([0, 5\pi]\)
The sine function is periodic with a period of \(2\pi\). Therefore, in each interval of \(2\pi\), we will have two solutions:
1. In the interval \([0, 2\pi]\): 2 solutions
2. In the interval \([2\pi, 4\pi]\): 2 solutions
3. In the interval \([4\pi, 5\pi]\): 2 solutions (only one complete cycle, but still two solutions)

Thus, the total number of solutions in the interval \([0, 5\pi]\) is:
\[ 2 + 2 + 2 = 6 \]

### Final Answer
The number of values of \(x\) in the interval \([0, 5\pi]\) satisfying the equation is \(6\).

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