bài 2
pt [TEX]\Leftrightarrow (\frac{1}{2}cos2x+\frac{\sqrt3}{2}sin2x)-5(\frac{1}{2}sinx+\frac{\sqrt3}{2}cosx)+4=0[/TEX]
[TEX]\Leftrightarrow cos(2x-\frac{\pi}{3})-5cos(x-\frac{\pi}{6})+4=0[/TEX]
[TEX]\Leftrightarrow cos2.(x-\frac{\pi}{6})-5cos(x-\frac{\pi}{6})+4=0[/TEX]
[TEX]\Leftrightarrow 2cos^2(x-\frac{\pi}{6})-1 -5cos(x-\frac{\pi}{6})+4=0[/TEX]
[TEX]\Leftrightarrow 2cos^2(x-\frac{\pi}{6})-5cos(x-\frac{\pi}{6})+3=0[/TEX]
[TEX]\Leftrightarrow cos(x-\frac{\pi}{6})=1[/TEX] hoặc [TEX]cos(x-\frac{\pi}{6})=\frac{3}{2}[/TEX](loại)
[TEX]\Rightarrow x-\frac{\pi}{6}=k2\pi[/TEX]
[TEX]\Leftrightarrow x=\frac{\pi}{6}+k2\pi[/TEX]