Ответы и объяснения

2012-02-27T17:30:12+04:00

\frac{6}{cos^2x}}-\frac{7}{cosx}+1=0

\frac{6-7cosx+cos^{2}x}{cos^{2}x}=0

cos^{2}x\neq0                         cos^{2}x-7cosx+6=0

x\neq\frac{pi}{2}+pin               cosx=t

                                                              t^{2}-7t+6=0

                                                              t=-1           t=\frac{1}{6}

                                             cosx=-1            cosx=\frac{1}{6}

                                x=pi+2pin         x=+-arccos\frac{1}{6}+2pin