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Question

Find all the roots of the root of the equation $$\displaystyle (1 \, - \, cos \, 2x) \, sin \, 2x \, = \, \sqrt{3} \, sin^2 \, x$$ which lie in the interval  $$\displaystyle \left [ 0, \, \frac{\pi}{3} \, \right ] .$$


Solution

$$\left(1-\cos{2x}\right)\sin{2x}=\sqrt{3}{\sin}^{2}{x}$$
$$\Rightarrow 2{\sin}^{2}{x}\sin{2x}=\sqrt{3}{\sin}^{2}{x}$$
$$\Rightarrow 2\sin{2x}=\sqrt{3}$$
$$\Rightarrow \sin{2x}=\dfrac{\sqrt{3}}{2}$$
$$\Rightarrow 2x=n\pi+\dfrac{\pi}{3}$$ or $$x=\dfrac{n\pi}{2}+\dfrac{\pi}{6}$$ is the general solution
The principal solution between $$\left[0,\dfrac{\pi}{3}\right]$$ is 
$$n=0\Rightarrow x=\dfrac{\pi}{6}$$
$$\therefore x=\dfrac{\pi}{6}$$ lies between $$\left[0,\dfrac{\pi}{3}\right]$$

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