Đáp án:
$S=\left\{\dfrac{1}{2}\arctan5+k\dfrac{\pi}{2}\,\bigg|\,k\in\mathbb Z\right\}$
Giải thích các bước giải:
ĐKXĐ: $\begin{cases}\cos x\ne 0\\\tan x\ne ±1\end{cases}⇔\begin{cases}x\ne\dfrac{\pi}{2}+k\pi\\x\ne±\dfrac{\pi}{4}+k\pi\end{cases}$
$\dfrac{2\tan x}{1-\tan^2x}=5$
$⇔\tan2x=5$
$⇒2x=\arctan5+k\pi\,\,(k\in\mathbb Z)$
$⇔x=\dfrac{1}{2}\arctan5+k\dfrac{\pi}{2}\,\,(k\in\mathbb Z)$
Vậy $S=\left\{\dfrac{1}{2}\arctan5+k\dfrac{\pi}{2}\,\bigg|\,k\in\mathbb Z\right\}$.