Find all solutions of each equation on the interval 0≤x< 2 pi.tan^2 x sec^2 x +2 sec^2 x - tan ^2 x= 2

Answer: [tex]x=0,\pi,2\pi[/tex]Step-by-step explanation:The given equation is: [tex]\tan^2x\sec^2x+2\sec^2x-\tan^2x=2[/tex].Subtract 2 from both sides[tex]\tan^2x\sec^x+2\sec^2x-\tan^2x-2=0[/tex].Factor by grouping:[tex]\sec^2x(\tan^2x+2)-1(\tan^2x+2)=0[/tex].[tex](\sec^2x-1)(\tan^2x+2)=0[/tex].Apply the zero product principle:[tex](\sec^2x-1)=0\:\:or\:\:(\tan^2x+2)=0[/tex].[tex]\sec^2x=1\:\:or\:\:\tan^2x=-2[/tex].If [tex]\sec^2x=1[/tex], then [tex]\sec x=\pm 1[/tex], [tex]\implies \cos x=\pm1[/tex]This implies that: [tex]x=0,\pi,2\pi[/tex]If [tex]\tan^2x=-2[/tex], x is not defined for all real values.Therefore the required solution on the given interval [tex]0\le x\le2\pi[/tex] is  [tex]x=0,\pi,2\pi[/tex]