Question

The number of solutions of the equation $2{\sin }^{2}x+\sqrt{3}\cos x+1=0$ in $[-\pi ,3\pi ]$ is

• A. $2$
• B. $3$
• C. $4$
• D. $6$

Answer 1

Answer: (C)

$4$

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

$\Rightarrow 2(1-{\cos }^{2}x)+\sqrt{3}\cos x+1=0$

$\Rightarrow 2-2{\cos }^{2}x+\sqrt{3}\cos x+1=0$

$\Rightarrow -2{\cos }^{2}x+\sqrt{3}\cos x+3=0$

$\Rightarrow 2{\cos }^{2}x-\sqrt{3}\cos x-3=0$

$\Rightarrow 2{\cos }^{2}x-2\sqrt{3}\cos x+\sqrt{3}\cos x-3=0$

$\Rightarrow 2\cos x(\cos x-\sqrt{3})+\sqrt{3}(\cos x-\sqrt{3})=0$

$\Rightarrow (\cos x-\sqrt{3})(2\cos x+\sqrt{3})=0$

$\Rightarrow \cos x\ne \sqrt{3}$ since range of $\cos x$ is

$[-1,1]$ and $\sqrt{3}=1.732>1$

$\therefore \cos x=\frac{-\sqrt{3}}{2}$

$\Rightarrow x=2n\pi \pm \frac{\pi }{6}$

$\Rightarrow x=\pm \frac{\pi }{6},2\pi \pm \frac{\pi }{6}$

$\therefore x=\{\frac{\pi }{6},\frac{-\pi }{6},\frac{11\pi }{6},\frac{13\pi }{6}\}$