Question

Number of roots of the equation ${\cos }^{2}x+\frac{\sqrt{3}+1}{2}\sin x-\frac{\sqrt{3}}{4}-1=0$ which lie in the interval $[-\pi ,\pi ]$ is-

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is B $4$

Given equation is ${\cos }^{2}x+\frac{\sqrt{3}+1}{2}\sin x-\frac{\sqrt{3}}{4}-1=0$

$\Rightarrow 1-{\sin }^{2}x+\frac{\sqrt{3}+1}{2}\sin x-\frac{\sqrt{3}}{4}-1=0$

$\Rightarrow {\sin }^{2}x-\frac{\sqrt{3}+1}{2}\sin x+\frac{\sqrt{3}}{4}=0$

$\Rightarrow 4{\sin }^{2}x-2\sqrt{3}\sin x-2\sin x+\sqrt{3}=0$
On solving we get $\sin x=\frac{1}{2}$;$\frac{\sqrt{3}}{2}$
Hence, $x=\frac{\pi }{6},\frac{5\pi }{6},\frac{\pi }{3},\frac{2\pi }{3}$