Đáp án:
`D`
Giải thích các bước giải:
`sin^2x-(sqrt3+1)sinxcosx+sqrt3cos^2x=sqrt3`
Xét `cosx=0=>` `sin^2x=1`
`PT<=>1=sqrt3(KTM)`
Xét `cosxne0`
Chia hai vế phương trình cho `cos^2x` ta được:
`tan^2x-(sqrt3+1)tanx+sqrt3=sqrt3(1+tan^2x)`
`<=>tan^2x-(sqrt3+1)tanx+sqrt3-sqrt3-sqrt3tan^2x=0`
`<=>(1-sqrt3)tan^2x-(sqrt3+1)tanx=0`
`<=>`\(\left[ \begin{array}{l}\tan x=-2-\sqrt3\\\tan x=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}\tan x=-2-\sqrt3\\\sin x=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}\tan x=-2-\sqrt3\\\sin^2x=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}\tan x=-2-\sqrt3\\1-\cos^2x=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}\tan x=-2-\sqrt3\\\cos^2x=1\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}\tan x +2+\sqrt3=0\\\cos^2x-1=0\end{array} \right.\)
`<=>(tan x +2+\sqrt3)(cos^2x-1)=0`
`toD`
