bài 2 [tex] 2cos(x+\frac{\pi}{6})= sin3x-cos3x[/tex]
đặt : [TEX]t = x + \frac{\pi}{6}[/TEX] \Rightarrow [TEX]3x = 3t - \frac{\pi}{2}[/TEX]
==>
[TEX]2 cos t = sin(3t - \frac{\pi}{2}) - cos (3t - \frac{\pi}{2})[/TEX]
[TEX]\Leftrightarrow 2 cos t = -cos 3t - sin 3t[/TEX]
[TEX]\Leftrightarrow2 cos t = -( 4 cos^3t - 3 cos t ) -(3 sint -4 sin^3t)[/TEX]
[TEX]\Leftrightarrow 4 cos^3 t - cost +3 sint - 4 sin^3t = 0[/TEX]
Xét 2 Th :
cost =0 -> ... ko là nghiệm
Cos t # o , chia 2 vế pt cho cost # 0 dc pt
......
[tex]tan^3 t + tan^2 t - 3 tant - 3 = 0[/tex]
[TEX]\Leftrightarrow(tant+1)(tan^2 t - 3) = 0[/TEX]
......