Question

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 1

`x=0,\pi,2\pi` on the interval
`0\le x\le 2\pi`

Explanation:

The given equation is:

`\tan^2x \sec^2x+2+2\sec^2x-\tan^2x=2`

We rearrange to get:

`\tan^2x \sec^2x+2+2\sec^2x-\tan^2x-2=0`

Factor by grouping:

`\sec^2x(\tan^2x+2)-1(\tan^2x+2)=0`

`(\sec^2x-1)(\tan^2x+2)=0`

Apply the zero product principle:

`(\sec^2x-1)=0\:\:\:Or\:\:\:(\tan^2x+2)=0`

When

`\sec^2x-1=0`

Then
`\sec x=\pm1`

This implies that:

`\cos x=\pm1`

`x=0,\pi,2\pi`

When
`\tan^2x+2+2=0`, we get
`\tan^2x=-2`, x is not defined for real values.