Giải thích các bước giải:
$VT=\sin\left (2x+\dfrac{\pi}{3} \right )\cos\left ( x-\dfrac{\pi}{6} \right )+\cos\left ( \dfrac{2\pi}{3}-2x \right )\cos\left ( \dfrac{2\pi}{3}-x \right )\\
=\left (\sin2x\cos\dfrac{\pi}{3}+\cos2x\sin\dfrac{\pi}{3} \right ).\left ( \cos x\cos\dfrac{\pi}{6}+\sin x\sin\dfrac{\pi}{6} \right )+\left ( \cos\dfrac{2\pi}{3}\cos2x+\sin\dfrac{2\pi}{3}\sin2x \right ).\left ( \cos\dfrac{2\pi}{3}\cos x+\sin\dfrac{2\pi}{3}\sin x \right )\\
=\left (\sin2x.\dfrac{1}{2}+\cos2x.\dfrac{\sqrt{3}}{2} \right ).\left ( \cos x.\dfrac{\sqrt{3}}{2}+\sin x.\dfrac{1}{2} \right )+\left ( \dfrac{-1}{2}.\cos2x+\dfrac{\sqrt{3}}{2}.\sin2x \right ).\left ( \dfrac{-1}{2}.\cos x+\dfrac{\sqrt{3}}{2}.\sin x \right )\\
=\dfrac{\sqrt{3}}{4}\sin2x.\cos x+\dfrac{1}{4}\sin x\sin2x+\dfrac{3}{4}\cos x\cos2x+\dfrac{\sqrt{3}}{4}\sin x\cos2x+\dfrac{1}{4}\cos x\cos2x-\dfrac{\sqrt{3}}{4}\sin x\cos2x-\dfrac{\sqrt{3}}{4}\cos x\sin2x+\dfrac{3}{4}\sin x\sin2x\\=\dfrac{1}{4}\sin x\sin2x+\dfrac{3}{4}\cos x\cos2x+\dfrac{1}{4}\cos x\cos2x+\dfrac{3}{4}\sin x\sin2x\\
=\dfrac{1}{4}\left ( \sin x\sin2x+\cos x\cos2x \right )+\dfrac{3}{4}\left ( \cos x\cos2x+\sin x\sin2x \right )\\
=\dfrac{1}{4}\cos(x-2x)+\dfrac{3}{4}\cos(x-2x)\\
=\dfrac{1}{4}\cos x+\dfrac{3}{4}\cos x\\
=\cos x=VP(đpcm)$
