Làm tam câu a trước vậy:Ta có: $ tanx-\frac{1}{tanx}+\frac{2cos4x}{sin2x}=0 $$ \Leftrightarrow tanx-\frac{1}{tanx}+\frac{2(2cos^22x-1)}{2sinx.cox}=0 $$ \Leftrightarrow tanx-\frac{1}{tanx}+\frac{4(2cos^2x-1)^2-2}{2tanx.cos^2x}=0 $$ \Leftrightarrow 2tan^2.cos^2x -18cos^2x +16cos^4x+2=0 $$ \Leftrightarrow 16cos^4x-20cos^2x+4=0 $ ( do $ tan^2x=\frac{1}{cos^2x}-1 $)suy ra $ cosx=\pm \frac{1}{2} $
Làm tam câu a trước vậy:Ta có: $ tanx-\frac{1}{tanx}+\frac{2cos4x}{sin2x}=0 $$ \Leftrightarrow tanx-\frac{1}{tanx}+\frac{2(2cos^22x-1)}{2sinx.cox}=0 $$ \Leftrightarrow tanx-\frac{1}{tanx}+\frac{4(2cos^2x-1)^2-2}{2tanx.cos^2x}=0 $$ \Leftrightarrow 2tan^2.cos^2x -18cos^2x +16cos^4x+2=0 $$ \Leftrightarrow 16cos^4x-20cos^2x+4=0 $ ( do $ tan^2x=\frac{1}{cos^2x}-1 $)suy ra $ cosx=\frac{1}{2} $
Làm tam câu a trước vậy:Ta có: $ tanx-\frac{1}{tanx}+\frac{2cos4x}{sin2x}=0 $$ \Leftrightarrow tanx-\frac{1}{tanx}+\frac{2(2cos^22x-1)}{2sinx.cox}=0 $$ \Leftrightarrow tanx-\frac{1}{tanx}+\frac{4(2cos^2x-1)^2-2}{2tanx.cos^2x}=0 $$ \Leftrightarrow 2tan^2.cos^2x -18cos^2x +16cos^4x+2=0 $$ \Leftrightarrow 16cos^4x-20cos^2x+4=0 $ ( do $ tan^2x=\frac{1}{cos^2x}-1 $)suy ra $ cosx=\
pm \frac{1}{2} $