Answer:[tex]x= 0[/tex] and [tex]x =\pi[/tex]Step-by-step explanation:Remember the following trigonometric property[tex]tan^2(x) + 1 = sec^2(x)[/tex]We have the following equation[tex]tan^2(x)*sec^2(x)+2sec^2(x)-tan^2(x) =2[/tex] for [tex]0\leq x <2\pi[/tex]Using the mentioned property we have to:[tex]tan^2(x)*(tan^2(x) + 1)+2(tan^2(x) + 1)-tan^2(x) =2[/tex][tex]tan^2(x)*(tan^2(x) + 1)+2tan^2(x) + 2-tan^2(x) =2[/tex][tex]tan^2(x)*(tan^2(x) + 1)+2tan^2(x) -tan^2(x) =0[/tex][tex]tan^2(x)*(tan^2(x) + 1)+tan^2(x) =0[/tex]Take [tex]tan^2(x)[/tex] as a common factor[tex]tan^2(x)*[(tan^2(x) + 1)+1] =0[/tex][tex]tan^2(x)*(tan^2(x) + 2) =0[/tex]Then:[tex]tan^2(x)= 0[/tex] or [tex]tan^2(x)+2=0[/tex] → [tex]tan^2(x)=-2[/tex][tex]tan(x) = 0[/tex] when [tex]x= 0[/tex] and [tex]x =\pi[/tex][tex]tan^2(x)=-2[/tex] there is no solution for this caseFinally the solutions are:[tex]x= 0[/tex] and [tex]x =\pi[/tex]