Question

If $4\sin^2 x - 2(1+\sqrt{3})\sin x + \sqrt{3} = 0$ and $15^\circ<x<150^\circ$, then the values of $x$ are:

• A. $30^\circ,45^\circ,90^\circ$
• B. $45^\circ,100^\circ,120^\circ$
• C. $30^\circ,60^\circ,120^\circ$
• D. $35^\circ,45^\circ,90^\circ$
• E. $30^\circ,45^\circ,130^\circ$

Hint:

Convert trigonometric equations into quadratic form whenever possible.

Answer 1

The Correct Option is C

Concept:
• Treat equation as quadratic in $\sin x$

Step 1: Let $\sin x = t$
\[ 4t^2 - 2(1+\sqrt{3})t + \sqrt{3} = 0 \]

Step 2: Solve quadratic
\[ t = \frac{2(1+\sqrt{3}) \pm \sqrt{[2(1+\sqrt{3})]^2 - 16\sqrt{3}}}{8} \] \[ = \frac{2(1+\sqrt{3}) \pm 2(1-\sqrt{3})}{8} \]

Step 3: Find roots
\[ t = \frac{4}{8} = \frac{1}{2}, \quad t = \frac{4\sqrt{3}}{8} = \frac{\sqrt{3}}{2} \]

Step 4: Find angles
For $\sin x = \frac{1}{2}$: \[ x = 30^\circ, 150^\circ \] For $\sin x = \frac{\sqrt{3}}{2}$: \[ x = 60^\circ, 120^\circ \]

Step 5: Apply interval $15^\circ<x<150^\circ$
Valid values: \[ 30^\circ, 60^\circ, 120^\circ \] Final Conclusion:
Option (C)