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 SOMEONE PLEASE HELPPP!!

Answer:[tex]x = 0 , \pi , 2\pi[/tex]Step-by-step explanation:The given equation is:l[tex] \tan^{2} (x) \sec^{2} x + 2 \sec^{2} x - \tan^{2} x = 2[/tex]Add -2 to both sides of the equation to get:[tex] \tan^{2} (x) \sec^{2} x + 2 \sec^{2} x - \tan^{2} x - 2 = 0[/tex]We factor the LHS by grouping.[tex]\sec^{2} x(\tan^{2} (x) + 2 ) - 1( \tan^{2} x + 2) = 0[/tex][tex](\sec^{2} x - 1)(\tan^{2} (x) + 2)= 0[/tex]We now apply the zero product property to get:[tex](\sec^{2} x - 1) = 0 \: \: or \: \: (\tan^{2} (x) + 2)= 0[/tex]This implies that:[tex]\sec^{2} x = 1 \: \: or \: \: \tan^{2} (x) = - 2[/tex][tex] \tan^{2} (x) = - 2 \implies \tan(x) = \pm \sqrt{ - 2} [/tex]This factor is never equal to zero and has no real solution.[tex]\sec^{2} x = 1[/tex]This implies that:[tex]\sec \: x= \pm\sqrt{1} [/tex][tex] \sec(x) = \pm - 1[/tex]Recall that[tex] \frac{1}{ \sec(x) } = \cos(x) [/tex]We reciprocate both sides to get:[tex] \cos(x) = \pm1[/tex][tex]\cos x = 1 \: or \: \cos x = - 1[/tex][tex]\cos x = 1 \implies \: x = 0 \: or \: 2\pi[/tex][tex]\cos x = - 1 \implies \: x = \pi[/tex]Therefore on the interval [tex]0 \leqslant x \leqslant 2\pi[/tex][tex]x = 0 , \pi , 2\pi[/tex]