Question

The solution set of the equation $4\sin \theta -2\cos \theta -2\sqrt{3}\sin \theta +\sqrt{3}=0$ in the interval $(0,2\pi )$ is-

• A. $\left\{\frac{3\pi }{4},\frac{7\pi }{4}\right\}$
• B. $\left\{\frac{\pi }{3},\frac{5\pi }{3}\right\}$
• C. $\left\{\frac{3\pi }{4},\frac{7\pi }{4},\frac{\pi }{3},\frac{5\pi }{3}\right\}$
• D. $\left\{\frac{\pi }{6},\frac{5\pi }{6},\frac{11\pi }{6}\right\}$

Answer 1

Answer: (D)

$\left\{\frac{\pi }{6},\frac{5\pi }{6},\frac{11\pi }{6}\right\}$

$4\sin \theta -2\cos \theta -2\sqrt{3}\sin \theta +\sqrt{3}=0$

$\Rightarrow (2\cos \theta -\sqrt{3})(2\sin \theta -1)=0$

$\therefore \cos \theta =\frac{\sqrt{3}}{2}$ or $\sin \theta =\frac{1}{2}$

$\theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{11\pi }{6}$

At $\theta =\frac{\pi }{6},\frac{11\pi }{6},\cos \theta =\frac{\sqrt{3}}{2}$

At $\theta =\frac{\pi }{6},\frac{5\pi }{6},\sin \theta =\frac{1}{2}$

$\therefore \theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{11\pi }{6}$