Question

Find all the roots of the root of the equation $(1-cos2x)sin2x=\sqrt{3}si{n}^{2}x$ which lie in the interval $[0,\frac{\pi }{3}].$

Answer 1

$(1-\cos 2x)\sin 2x=\sqrt{3}{\sin }^{2}x$

$\Rightarrow 2{\sin }^{2}x\sin 2x=\sqrt{3}{\sin }^{2}x$

$\Rightarrow 2\sin 2x=\sqrt{3}$

$\Rightarrow \sin 2x=\frac{\sqrt{3}}{2}$

$\Rightarrow 2x=n\pi +\frac{\pi }{3}$ or $x=\frac{n\pi }{2}+\frac{\pi }{6}$ is the general solution

The principal solution between $[0,\frac{\pi }{3}]$ is

$n=0\Rightarrow x=\frac{\pi }{6}$

$\therefore x=\frac{\pi }{6}$ lies between $[0,\frac{\pi }{3}]$