Question

If $4{\sin }^2\theta +2(\sqrt{3}+1)\cos \theta =4+\sqrt{3}$, then the general solution is

• A. $n\pi \pm \frac{\pi }{3},n\in Z$
• B. $n\pi \pm \frac{\pi }{6},n\in Z$
• C. $2n\pi \pm \frac{\pi }{3},n\in Z$
• D. $2n\pi \pm \frac{\pi }{4},n\in Z$

Answer 1

The correct option is C $2n\pi \pm \frac{\pi }{3},n\in Z$

Given: $4{\sin }^2\theta +2(\sqrt{3}+1)\cos \theta =4+\sqrt{3}$
$\Rightarrow 4-4{\cos }^2\theta +2(\sqrt{3}+1)\cos \theta =4+\sqrt{3}\Rightarrow 4{\cos }^2\theta -2(\sqrt{3}+1)\cos \theta +\sqrt{3}=0\Rightarrow 4{\cos }^2\theta -2\cos \theta -2\sqrt{3}\cos \theta +\sqrt{3}=0\Rightarrow (2\cos \theta -\sqrt{3})(2\cos \theta -1)=0\Rightarrow (2\cos \theta -\sqrt{3})(2\cos \theta -1)=0\Rightarrow \cos \theta =\frac{\sqrt{3}}{2}\text{or}\cos \theta =\frac{1}{2}\therefore \theta =2n\pi \pm \frac{\pi }{6}\text{or}\theta =2n\pi \pm \frac{\pi }{3},n\in Z$