ĐK tự xử nhé
$\frac{2sinx}{2cos2x - 1} + \frac{1}{2sinx + 1} = \frac{1}{2} + 2cos(x + \frac{\pi }{3})cos(x - \frac{\pi }{3})$
$\Leftrightarrow \frac{2sinx}{1-4\sin^2 x} + \frac{1}{2sinx + 1} = \frac{1}{2} + 2cos(x + \frac{\pi }{3})cos(x - \frac{\pi }{3})$
$\Leftrightarrow \frac{2sinx +1-2\sin x}{1-4\sin^2 x} = \frac{1}{2} + 2cos(x + \frac{\pi }{3})cos(x - \frac{\pi }{3})$
$\Leftrightarrow \dfrac{1}{2\cos 2x -1} = \frac{1}{2} +\cos^2 x -\sin^2 x -\frac{1}{2}$
$\Leftrightarrow \dfrac{1}{2\cos 2x -1}=\cos 2x$ Quá dễ rồi nhé