Đáp án:
$VT=4sin(x+\frac{\pi}{3})sin(x-\frac{\pi}{3})\\
=4.\frac{1}{2}.\left [ cos(x+\frac{\pi}{3}-x+\frac{\pi}{3})-cos(x+\frac{\pi}{3}+x-\frac{\pi}{3}) \right ]\\
=2(cos\frac{2\pi}{3}-cos2x)\\
=2(\frac{-1}{2}-cos2x)\\
=-1-2cos2x\\
=-1-2(2cos^2x-1)\\
=-1-4cos^2x+2\\
=1-4cos^2x=VP(ĐPCM)$