[tex]-4-\sqrt{3}cosx+cos(2x)+sinx-\sqrt{3}sin(2x)=0\Leftrightarrow -4+(sinx-\sqrt{3}cosx)+[cos(2x)-\sqrt{3}sin(2x)]=0\Leftrightarrow -4+2sin(\frac{\pi}{6}-2x)-2cos(x+\frac{\pi}{6})=0\Leftrightarrow sin(\frac{\pi}{6}-2x)-cos(x+\frac{\pi}{6})-2=0[/tex]
Vì [TEX]sin(\frac{\pi}{6}-2x) \leq 1, cos(x+\frac{\pi}{6}) \geq -1[/TEX] nên [TEX]VT \leq 0[/TEX]
Dấu "=" xảy ra khi [tex]\Leftrightarrow \left\{\begin{matrix} \frac{\pi}{6}-2x=\frac{\pi}{2}+2k\pi\\ x+\frac{\pi}{6}=2k\pi+\pi \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=-\frac{\pi}{6}-k\pi\\ x=2k\pi+\frac{5\pi}{6} \end{matrix}\right.\Leftrightarrow x=2k\pi+\frac{5\pi}{6}[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
