Đáp án:
Giải thích các bước giải:
`5cos\ 2x+sin^2x+2cosx+1=0.`
`⇔ 5(2cos^2 x-1)+(1-cos^2 x)+2cos\ x+1=0`
`⇔ 10cos^2 x-5+1-cos^2 x+2cos\ x+1=0`
`⇔ 9cos^2 x+2cos\ x-3=0`
`⇔` \(\left[ \begin{array}{l}cos\ x=\dfrac{-1+2\sqrt{7}}{9}\\cos\ x=\dfrac{-1-2\sqrt{7}}{9}\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=arccos (\dfrac{-1+2\sqrt{7}}{9})+k2\pi\ (k \in \mathbb{Z})\\x=-arccos (\dfrac{-1+2\sqrt{7}}{9})+k2\pi\ (k \in \mathbb{Z})\\x=arccos (\dfrac{-1-2\sqrt{7}}{9})+k2\pi\ (k \in \mathbb{Z})\\x=-arccos (\dfrac{-1-2\sqrt{7}}{9})+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
Vậy `S={arccos (\frac{-1+2\sqrt{7}}{9})+k2\pi\ (k \in \mathbb{Z});-arccos (\frac{-1+2\sqrt{7}}{9})+k2\pi\ (k \in \mathbb{Z});arccos (\frac{-1-2\sqrt{7}}{9})+k2\pi\ (k \in \mathbb{Z});-arccos (\frac{-1-2\sqrt{7}}{9})+k2\pi\ (k \in \mathbb{Z})}`
