Giải thích các bước giải:
Ta có:
$4\sin(x+\dfrac{\pi}{3})\cdot \sin(x-\dfrac{\pi}{3})=4\sin^2x-3$
$\to 4\cdot (\sin x\cos\dfrac{\pi}{3}+\cos x\sin\dfrac{\pi}{3})\cdot (\sin x\cos\dfrac{\pi}{3}-\cos x\sin\dfrac{\pi}{3})=4\sin^2x-3$
$\to 4\cdot (\sin x\cdot\dfrac12+\cos x\cdot \dfrac{\sqrt{3}}{2})\cdot (\sin x\cdot\dfrac12-\cos x\cdot \dfrac{\sqrt{3}}{2})=4\sin^2x-3$
$\to (\sin x+\cos x\cdot \sqrt{3})\cdot (\sin x-\cos x\cdot \sqrt{3})=4\sin^2x-3$
$\to \sin^2x-3\cos^2x=4\sin^2x-3$
$\to 4\sin^2x-3(\sin^2x+\cos^2x)=4\sin^2x-3$
$\to 4\sin^2x-3\cdot 1=4\sin^2x-3$
$\to 4\sin^2x-3=4\sin^2x-3$ đúng
$\to đpcm$
