Đáp án:
Giải thích các bước giải:
`4cos 2x+3=0`
`⇔ 4cos 2x=-3`
`⇔ cos 2x=- 3/4`
`⇔ cos 2x = arccos (- 3/4)`
`⇔` \(\left[ \begin{array}{l}2x=arccos (-\dfrac{3}{4})+k2\pi\ (k \in \mathbb{Z})\\2x=-arccos (-\dfrac{3}{4})+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{arccos (-\dfrac{3}{4})}{2}+k\pi\ (k \in \mathbb{Z})\\x=\dfrac{-arccos (-\dfrac{3}{4})}{2}+k\pi\ (k \in \mathbb{Z})\end{array} \right.\)
Vậy `S={\frac{arccos (-\frac{3}{4})}{2}+k\pi\ (k \in \mathbb{Z}); \frac{-arccos (-\frac{3}{4})}{2}+k\pi}`
