Đáp án + Giải thích các bước giải:
`cos2x+sqrt3sin2x=-sqrt3`
`<=>1/2cos2x+sqrt3/2sin2x=-sqrt3/2`
`<=>sinfrac{pi}{6}cos2x+cosfrac{pi}{6}sin2x=-sqrt3/2`
`<=>sin(pi/6+2x)=sin(-pi/3)`
`<=>`\(\left[ \begin{array}{l}\dfrac{\pi}{6}+2x=-\dfrac{\pi}{3}+k2\pi\\\dfrac{\pi}{6}+2x=\pi+\dfrac{\pi}{3}+k2\pi\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}2x=-\dfrac{\pi}{2}+k2\pi\\2x=\dfrac{7\pi}{6}+k2\pi\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-\dfrac{\pi}{4}+k\pi\\x=\dfrac{7\pi}{12}+k\pi\end{array} \right.\)`(kinZZ)`
