ĐK: $x\ne \dfrac{\pi}{2}+k\pi$
Hoành độ giao:
$\sin2x=\tan x$
$\Leftrightarrow \sin2x.\cos x=\sin x$
$\Leftrightarrow 2\sin x.\cos^2x-\sin x=0$
$\Leftrightarrow \sin x(2\cos^2x-1)=0$
$\Leftrightarrow \sin x=0$ hoặc $\cos^2x=\dfrac{1}{2}$
$+) \sin x=0\Leftrightarrow x=k\pi$ (TM)
$\cos^2x=\dfrac{1+\cos2x}{2}=\dfrac{1}{2}\Leftrightarrow \cos2x=0\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}$ (TM)
$x=k\pi\Rightarrow y=\tan(k\pi)=0$
$x=\dfrac{\pi}{4}+k\dfrac{k\pi}{2}\Rightarrow y=\tan(\dfrac{\pi}{4}+\dfrac{k\pi}{2})=1$
Vậy các giao điểm có toạ độ $(k\pi;0)$, $(\dfrac{\pi}{4}+\dfrac{k\pi}{2};1)$ ($k\in\mathbb{Z}$)
