Question

The equation $4si{n}^{2}x-2(\sqrt{3}+1)sinx+\sqrt{3}=0$ has

• A. $2$ solutions in $(0,\pi )$
• B. $4$ solutions in $(0,2\pi )$
• C. $2$ solutions in $(-\pi ,\pi )$
• D. $4$ solutions in $(-\pi ,\pi )$

Answer 1

Answer: (B), (D)

$4$ solutions in $(0,2\pi )$
$4$ solutions in $(-\pi ,\pi )$

Put $\sin x=t$ in the given equation, we get

$4t^2-2(\sqrt{3}+1)t+\sqrt{3}=02t(2t-\sqrt{3})-1(2t-\sqrt{3})=0(2t-\sqrt{3})(2t-1)=0$

$t=\frac{1}{2}$ or $\frac{\sqrt{3}}{2}$

$\sin x=\frac{1}{2}$ or $\frac{\sqrt{3}}{2}$

$x=n\pi +(-1{)}^n\frac{\pi }{6}$ or $x=n\pi +(-1{)}^n\frac{\pi }{3}$

$x=\frac{\pi }{3},\frac{\pi }{6},\frac{5\pi }{6}$ and $\frac{2\pi }{3}$

Therefore the above equation has $4$ solutions in $(0,2\pi ),(0,\pi )$ and $(-\pi ,\pi )$