[TEX](sinx + \sqrt{3}cosx)sin3x = 2[/TEX]
[TEX]\Leftrightarrow 2sin(x+\frac{\pi}{3})sin3x = 2[/TEX]
[TEX]\Leftrightarrow sin(x+\frac{\pi}{3})sin3x = 1[/TEX]
Vì [TEX] |sin(x+ \frac{\pi}{3}| \leq 1 [/TEX]
và [TEX] |sinx| \leq 1[/TEX]
Suy ra:
TH1:
[TEX]\left{\begin{sin(x+\frac{\pi}{3})=1}\\{sin3x =1}[/TEX]
\Leftrightarrow [TEX]\left{\begin{x+\frac{\pi}{3}=\frac{\pi}{2}+ k2\pi}\\{3x =\frac{\pi}{2}+ k2\pi} \Leftrightarrow ...[/TEX]
TH2:
[TEX] \left{\begin{sin(x+\frac{\pi}{3})=-1}\\{sin3x =-1}[/TEX]
[TEX] \Leftrightarrow \left{\begin{x+\frac{\pi}{3}=\frac{-\pi}{2}+ k2\pi}\\{3x =\frac{-\pi}{2}+ k2\pi}[/TEX]
Sau đó kết hợp cả 2 TH (k chắc đúng k)