Find all solutions of each equation on the interval [tex]0\leq x\ \textless \ 2\pi .[/tex][tex]tan^{2} x[/tex][tex]sec^{2} x[/tex] + 2 [tex]sec^{2}x[/tex][tex]-tan^{2} x[/tex] =2

Answer:[tex]x_1=0\\ \\x_2=\pi[/tex]Step-by-step explanation:Use definitions of [tex]\tan x[/tex] and [tex]\sec x:[/tex][tex]\tan x=\dfrac{\sin x}{\cos x}\\ \\\sec  x=\dfrac{1}{\cos x}[/tex]Hence, the equation is[tex]\left(\dfrac{\sin x}{\cos x}\right)^2\cdot \left(\dfrac{1}{\cos x}\right)^2+2\left(\dfrac{1}{\cos x}\right)^2-\left(\dfrac{\sin x}{\cos x}\right)^2=2[/tex]Multiply this equation by [tex]\cos ^4x:[/tex][tex]\sin^2 x+2\cos^2 x-\sin^2x\cos^2x=2\cos^4x\\ \\\sin^2 x+2\cos^2 x-\sin^2x\cos^2x-2\cos^4x=0\\ \\\sin^2x(1-\cos^2x)+2\cos^2x(1-\cos^2x)=0\\ \\(1-\cos^2x)(\sin^2x+2\cos^2x)=0[/tex]This means that[tex]1-cos^2x=0\ \text{or} \ \sin^2x+2\cos^2x=0[/tex]The second equation has no solutions, because the sum of two non-negative numbers is 0 only if both these numbers are 0 and this is impossible, because [tex]\sin x[/tex] and [tex]\cos x[/tex] cannot be 0 simultaneously.Solve the first equation:[tex]\cos^2x=1\Rightarrow \\ \\\cos x=1\ \text{or}\ \cos x=-1\\ \\x=2\pi k,\ k\in Z\ \ \text{or}\ \ x=\pi+2\pi k,\ k\in Z[/tex]If [tex]0\le x<2\pi,[/tex] then[tex]x_1=0\\ \\x_2=\pi[/tex]