Question 979394
{{{sec^2(x)-6sec(x)-7=0}}}--->{{{(sec(x)-7)(sec(x)+1)=0}}}
(factoring just like you would do {{{u^2-6u-7=0}}}--->{{{(u-7)(u+1)=0}}} )

{{{(sec(x)-7)(sec(x)+1)=0}}}--->{{{system(sec(x)-7=0,"or",sec(x)+1=0)}}}--->{{{system(sec(x)=7,"or",sec(x)=-1)}}}
So, option {{{highlight(A)}}} lists two solutions.