Đáp án:
1) $\left[ \begin{array}{l}x=\dfrac{\pi}{ 4} +\dfrac{k\pi}{ 2}\\ x =k\pi\end{array} \right. (k\in\mathbb{Z})$
2) $\left[ \begin{array}{l}x=\dfrac{k\pi}{2}\\x=\dfrac{\pi}{2}+k\pi\end{array} \right. (k\in\mathbb{Z})$
Giải thích các bước giải:
1) $\cos 2x . \tan x =0\ (*)$
ĐKXĐ: $x\neq \dfrac{\pi}{ 2} + k\pi$
$(*)⇔\left[ \begin{array}{l}\cos 2x=0\\\sin x =0\end{array} \right.$
$⇔\left[ \begin{array}{l}2x=\dfrac{\pi}{ 2} + k\pi\\x =k\pi\end{array} \right.$
$⇔\left[ \begin{array}{l}x=\dfrac{\pi}{ 4} +\dfrac{k\pi}{ 2}\\ x =k\pi\end{array} \right. (k\in\mathbb{Z})$
2) $\sin 2x . \cot x =0\ (*)$
ĐKXĐ: $x\neq k\pi$
$(*)⇔\left[ \begin{array}{l}\sin 2x=0\\\cos x=0\end{array} \right.$
$⇔\left[ \begin{array}{l}2x=k\pi\\x=\dfrac{\pi}{2}+k\pi\end{array} \right.$
$⇔\left[ \begin{array}{l}x=\dfrac{k\pi}{2}\\x=\dfrac{\pi}{2}+k\pi\end{array} \right. (k\in\mathbb{Z})$
