Điều kiện: [tex]x \neq \frac{k\Pi }{2}[/tex]
[tex]\frac{1+cos2x}{cosx} = \frac{sin2x}{1-cos2x}[/tex]
[tex]\Leftrightarrow \frac{sin^2x + cos^2x +cos^2x - sin^2x}{cosx} = \frac{2sinxcosx}{1-1+2sin^2x}[/tex]
[tex]\Leftrightarrow \frac{2cos^2x}{cosx} = \frac{cosx}{sinx}[/tex] [tex]\Leftrightarrow \frac{1}{sinx} = 2[/tex]
[tex]\Leftrightarrow sinx = \frac{1}{2}[/tex]
[tex]\frac{\pi}{6} + \frac{5\pi}{6} = \pi[/tex]
[tex]sinxcosx(1+tanx)(1+cotx) = 1[/tex]
[tex]\Leftrightarrow (sinx + cosx)(sinx + cosx) = 1[/tex]
[tex]\Leftrightarrow sin^2x + cos^2x + 2sinxcosx = 1[/tex]
[tex]\Leftrightarrow 2sinxcosx = 0[/tex]
[tex]sinx = 0; cosx = 0[/tex]
[tex]x = k\pi; x = \frac{\pi}{2} + k\pi[/tex]
[tex]x = \frac{k\pi}{2}[/tex]
[tex]sin^2x + sin^23x = cos^2x + cos^3x[/tex]
[tex]\Leftrightarrow -cos2x = cos6x[/tex]
[tex]\Leftrightarrow cos2x + cos6x = 0[/tex]
[tex]\Leftrightarrow 2cos4xcos2x = 0[/tex]
[tex]\Leftrightarrow cos4x = 0; cos2x = 0[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
