Bài 2:Điều kiện[TEX]cosx\not=0\[/TEX]
cos2x + cosx($2tan^2 x$-1)=2
\Leftrightarrowcos2xcosx+$2sin^2x$-$cos^2x$-2cosx=0
\Leftrightarrow(2$cos^2x$-1)cosx+2(1-$cos^2x$)-$cos^2x$-2cosx=0
\Leftrightarrow2$cox^3x$-cosx-2$cos^2x$+2-$cos^2x$-2cosx=0
\Leftrightarrow2$cos^3x$-3$cos^2x$-3cosx+2=0
\Leftrightarrow[TEX]\left[\begin{cosx=-1}\\{cosx=1/2}[/TEX] (nghiệm cosx=2 loại)
cosx=-1\Leftrightarrowx=-[TEX]\pi[/TEX]+k2[TEX]\pi[/TEX]
cosx=1/2\Leftrightarrowx=(+-)[TEX]\pi/3[/TEX]+k2[TEX]\pi[/TEX] (k[TEX]\in[/TEX]Z)
(Thoã mãn điều kiện)