If number of solution and sum of solution of the equation ` 3sin^(2)x-7sinx +2=0, x in [0,2pi]` are respectively N and S and ` f_(n)(theta)=sin^(n)theta +cos^(n) theta`. On the basis of above information , answer the following questions.
Value of S is

Options

A. `(5pi)/(6)`
B. `(7pi)/(6)`
C. `2pi`
D. `pi`

Answer

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

### Step 1: Rewrite the equation
We start with the quadratic equation:
\[ 3\sin^2 x - 7\sin x + 2 = 0 \]
Let \( y = \sin x \). The equation becomes:
\[ 3y^2 - 7y + 2 = 0 \]

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

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

### Step 4: Determine valid solutions
Since \( y = \sin x \), we know that \( \sin x \) must be in the range \([-1, 1]\). Therefore, \( y = 2 \) is not a valid solution. We only consider:
\[ \sin x = \frac{1}{3} \]

### Step 5: Find the angles
To find the angles \( x \) where \( \sin x = \frac{1}{3} \) in the interval \( [0, 2\pi] \):
- The first solution is \( x_1 = \arcsin\left(\frac{1}{3}\right) \).
- The second solution in the interval is \( x_2 = \pi - \arcsin\left(\frac{1}{3}\right) \).

### Step 6: Calculate the sum of solutions
The sum of the solutions \( S \) is:
\[ S = x_1 + x_2 = \arcsin\left(\frac{1}{3}\right) + \left(\pi - \arcsin\left(\frac{1}{3}\right)\right) = \pi \]

### Conclusion
Thus, the value of \( S \) is:
\[ \boxed{\pi} \]