Question

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

• A. 4
• B. 6
• C. 2
• D. 1

Hint:

Check the given range $[0, 3\pi]$ carefully; it covers one and a half full rotations of the unit circle.

Answer 1

The Correct Option is A

Step 1: Solve the Quadratic
Let $\sin x = t$. Equation becomes $2t^2 + 5t - 3 = 0$.
$2t^2 + 6t - t - 3 = 0 \implies 2t(t+3) - 1(t+3) = 0$.
So, $t = 1/2$ or $t = -3$.
Step 2: Filter Valid Values
Since $|\sin x| \le 1$, $t = -3$ is rejected.
We solve $\sin x = 1/2$ for $x \in [0, 3\pi]$.
Step 3: Identify Solutions
In $[0, 2\pi]$: $x = \pi/6, 5\pi/6$.
In $[2\pi, 3\pi]$: $x = 2\pi + \pi/6 = 13\pi/6$ and $x = 2\pi + 5\pi/6 = 17\pi/6$.
Step 4: Conclusion
There are 4 solutions in total.
Final Answer:(A)