Question

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

The correct option is C $\left\{\frac{3\pi }{4},\frac{7\pi }{4},\frac{\pi }{3},\frac{5\pi }{3}\right\}$

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

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

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

$\sin \theta =\frac{1}{2},$

$\cos \theta =\frac{\sqrt{3}}{2},$

$\sin \theta =\sin \frac{\pi }{6}$

$\theta =n\pi +{\left(-1\right)}^n\frac{\pi }{6}$

$\theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{13\pi }{6},\frac{17\pi }{6}$

$\cos \theta =\cos \frac{\pi }{6}$

$\theta =2n\pi \pm \frac{\pi }{6}$

$\theta =\frac{\pi }{6},\frac{11\pi }{6},\frac{13\pi }{6}$

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