Đáp án: $x\in\{\dfrac13(\pi+\dfrac{u}{3}+k2\pi),\dfrac{u}{3}-k2\pi\}$
Giải thích các bước giải:
Ta có:
$\sin x=\sin(2x-\dfrac{u}{3})$
$\to x=2x-\dfrac{u}{3}+k2\pi$
$\to x=\dfrac{u}{3}-k2\pi$
Hoặc $x=\pi-(2x-\dfrac{u}{3})+k2\pi$
$\to x=\pi-2x+\dfrac{u}{3}+k2\pi$
$\to 3x=\pi+\dfrac{u}{3}+k2\pi$
$\to x=\dfrac13(\pi+\dfrac{u}{3}+k2\pi)$