Đáp án + Giải thích các bước giải:
$y=\dfrac{1+cosx}{cos2x.sin4x}$
ĐKXĐ: $cos2x.sin4x\neq0$
⇔ \(\left[ \begin{array}{l}cos2x\neq0\\sin4x\neq0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}2x\neq \dfrac{\pi}{2}+k\pi\\4x\neq k\pi\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x\neq \dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x\neq \dfrac{k\pi}{4}\end{array} \right.\) $(k∈\mathbb{Z})$
TXĐ: $D=R$ \ $\{\dfrac{\pi}{2}+\dfrac{k\pi}{2},k∈\mathbb{Z}\}$ ∪ $\{\dfrac{k\pi}{4},k∈\mathbb{Z}\}$
