Answer:
[tex] x=0,\pi,2\pi[/tex] on the interval [tex]0\le x\le 2\pi[/tex]
Step-by-step explanation:
The given equation is:
[tex]\tan^2x \sec^2x+2+2\sec^2x-\tan^2x=2[/tex]
We rearrange to get:
[tex]\tan^2x \sec^2x+2+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]
When
[tex] \sec^2x-1=0[/tex]
Then [tex] \sec x=\pm1[/tex]
This implies that:
[tex] \cos x=\pm1[/tex]
[tex] x=0,\pi,2\pi[/tex]
When [tex]\tan^2x+2+2=0[/tex], we get [tex]\tan^2x=-2[/tex], x is not defined for real values.
