Step 1: Understanding the Question:
We need to find the number of real solutions for the given quadratic equation in \( \sin x \) within the domain \( [0, 3\pi] \).
Step 2: Key Formula or Approach:
Let \( t = \sin x \). The equation becomes a standard quadratic equation \( 2t^2 + 5t - 3 = 0 \). Solve for \( t \) and then find the corresponding values of \( x \).
Step 3: Detailed Explanation:
Solving \( 2t^2 + 5t - 3 = 0 \):
\[ 2t^2 + 6t - t - 3 = 0 \implies 2t(t + 3) - 1(t + 3) = 0 \]
\[ (2t - 1)(t + 3) = 0 \implies t = \frac{1}{2} \text{ or } t = -3 \]
Since \( \sin x \) must lie in the range \( [-1, 1] \), \( \sin x = -3 \) is not possible.
Thus, \( \sin x = \frac{1}{2} \).
In the interval \( [0, 2\pi] \), \( \sin x = \frac{1}{2} \) at \( x = \frac{\pi}{6} \) and \( x = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \).
In the interval \( [2\pi, 3\pi] \), \( \sin x = \frac{1}{2} \) at \( x = 2\pi + \frac{\pi}{6} = \frac{13\pi}{6} \) and \( x = 3\pi - \frac{\pi}{6} = \frac{17\pi}{6} \).
All these values lie within \( [0, 3\pi] \).
Total number of solutions = 4.
Step 4: Final Answer:
There are 4 solutions.