Question

Solve $2{\sin }^{2}x+\sqrt{3}\cos x+1=0$.

Answer 1

Changing ${\sin }^{2}x$ into $1-{\cos }^{2}x$; we get
$2(1-{\cos }^{2}x)+\sqrt{3}\cos x+1=0$
or $2{\cos }^{2}x-\sqrt{3}\cos x-3=0$
$\therefore \cos x=\frac{\sqrt{3}\pm \sqrt{3+24}}{4}$
$=\frac{\sqrt{3}+3\sqrt{3}}{4}=\sqrt{3}$ or $\frac{-\sqrt{3}}{2}$
Since $\sqrt{3}$ is greater than $1$ it is not admissible as $\cos x$ can not be greater than $1$.
$\therefore \cos x=-\sqrt{3}/2=-\cos \left(\pi /6\right)$
$=\cos (\pi -\pi /6)=\cos \left(5\pi /6\right)$
$\therefore x=2n\pi \pm \left(5\pi /6\right)$.