Đáp án:
a) $S=\left\{\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\bigg{|}\,k\in\mathbb Z\right\}$
b) $S=\emptyset$
Giải thích các bước giải:
a) ĐKXĐ: $x\ne \dfrac{k\pi}{2}$
$\cot2x=\tan x$
$⇔\tan x=\tan\left(\dfrac{\pi}{2}-2x\right)$
$⇔x=\dfrac{\pi}{2}-2x+k2\pi\,\,(k\in\mathbb Z)$
$⇔3x=\dfrac{\pi}{2}+k2\pi\,\,(k\in\mathbb Z)$
$⇔x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\,\,(k\in\mathbb Z)$
Vậy $S=\left\{\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\bigg{|}\,k\in\mathbb Z\right\}$
b) ĐKXĐ: $\begin{cases} x\ne k\pi\\x\ne -\dfrac{\pi}{6}-k\pi\end{cases}$
$\tan\left(\dfrac{\pi}{3}-x\right)=\cot x$
$⇔\tan\left(\dfrac{\pi}{3}-x\right)=\tan\left(\dfrac{\pi}{2}-x\right)$
$⇔\dfrac{\pi}{3}-x=\dfrac{\pi}{2}-x+k\pi\,\,(k\in\mathbb Z)$
$⇔-\dfrac{\pi}{6}=k\pi\,\,(k\in\mathbb Z)$
$⇒$ Phương trình vô nghiệm
Vậy $S=\emptyset$.
