5.
[tex]DK:...\\tanx+tan ( x+\frac{\pi}{3} )+tan ( x+\frac{2\pi}{3} \ )=3\sqrt{3}\\\Leftrightarrow \frac{sinx}{cosx}+\frac{sin( x+\frac{\pi}{3} ).cos( x+\frac{2\pi}{3} )+sin( x+\frac{2\pi}{3} ).cos( x+\frac{\pi}{3})}{cos( x+\frac{\pi}{3} )cos( x+\frac{2\pi}{3} )}=3\sqrt{3}\\\Leftrightarrow \frac{sinx}{cosx}+\frac{sin(2x+\pi)}{\frac{1}{2}(cos(2x+\pi)+cos(\frac{\pi}{3}))}=3\sqrt{3}\\\Leftrightarrow \frac{sinx}{cosx}-\frac{4sin2x}{1-2cos2x}=3\sqrt{3}\\\Leftrightarrow \frac{sinx-2sinxcos2x-4sin2xcosx}{cosx-2cosxcos2x}=3\sqrt{3}\\\Leftrightarrow \frac{sinx-sin3x+sinx-2sin3x-2sinx}{cosx-cosx-cos3x}=3\sqrt{3}\\\Leftrightarrow 3tan3x=3\sqrt{3}\\\Leftrightarrow ...[/tex]
Không biết còn cách nào hay hơn không chứ pha xử lí của mình hơi cồng kềnh :v
4.
$DK:...\\8cot2x=\frac{(cos^{2}x-sin^{2}x).sin2x}{cos^{6}x+sin^{6}x}\\\Leftrightarrow 8\frac{cos2x}{sin2x}=\frac{cos2x.sin2x}{cos^6x+sin^6x}$
Với $cos2x=0\\\Leftrightarrow...$
Với
$8(sin^6x+cos^6x)=sin^22x$
$\Leftrightarrow 8(sin^2x+cos^2x)(sin^4x-sin^2xcos^2x+cos^4x)=sin^22x$
$\Leftrightarrow 8sin^4x+8cos^4x-3sin^22x=0$
$\Leftrightarrow 8sin^4x+16sin^2xcos^2x+8cos^4x-12sin^2xcos^2x-16sin^2xcos^2x=0$
$\Leftrightarrow 28sin^2xcos^2x=8$
$\Leftrightarrow 7sin^22x=8$
$\Leftrightarrow sin^22x=\frac{8}{7} \leq 1 (VL)$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
