Question

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

• A. 0
• B. 5
• C. 6
• D. 10

Hint:

For trigonometric equations, solve the equivalent quadratic equation, and then use the inverse sine function to find the values of \( x \) in the given interval.

Answer 1

The Correct Option is C

Step 1: Solve the given trigonometric equation.
We are given the equation \( 3\sin^2 x - 7\sin x + 2 = 0 \).
Let \( y = \sin x \), so the equation becomes: \[ 3y^2 - 7y + 2 = 0 \]

Step 2: Solve the quadratic equation.
To solve for \( y \), use the quadratic formula: \[ y = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(3)(2)}}{2(3)} = \frac{7 \pm \sqrt{49 - 24}}{6} = \frac{7 \pm \sqrt{25}}{6} = \frac{7 \pm 5}{6} \] Thus, we have two possible values for \( y \): \[ y = \frac{7 + 5}{6} = 2 \quad \text{or} \quad y = \frac{7 - 5}{6} = \frac{1}{3} \]

Step 3: Analyze the solutions for \( y \).
Recall that \( y = \sin x \). Therefore: - For \( \sin x = 2 \), there are no solutions since \( \sin x \) cannot be greater than 1. - For \( \sin x = \frac{1}{3} \), we solve for \( x \).

Step 4: Find the values of \( x \).
For \( \sin x = \frac{1}{3} \), the general solutions are: \[ x = \sin^{-1}\left(\frac{1}{3}\right) + 2n\pi \quad \text{or} \quad x = \pi - \sin^{-1}\left(\frac{1}{3}\right) + 2n\pi \] Thus, the two solutions within each period \( [0, 2\pi] \) are \( x_1 \) and \( x_2 \).

Step 5: Count the number of solutions in the interval \( [0, 5\pi] \).
Since the period of the sine function is \( 2\pi \), there are 3 full periods in the interval \( [0, 5\pi] \), so we get 6 solutions.

Step 6: Conclusion.
Thus, the number of values of \( x \) that satisfy the equation is 6, corresponding to option (C).