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]
