Question

Find all solutions of each equation on the interval 0 ≤ x < 2 pi.

Answer 1

Answer:

$x=0,\pi,2\pi$

Step-by-step explanation:

The given equation is: $\tan^2x\sec^2x+2\sec^2x-\tan^2x=2$.

Subtract 2 from both sides

$\tan^2x\sec^x+2\sec^2x-\tan^2x-2=0$.

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:

$(\sec^2x-1)=0\:\:or\:\:(\tan^2x+2)=0$.

$\sec^2x=1\:\:or\:\:\tan^2x=-2$.

If $\sec^2x=1$, then $\sec x=\pm 1$,

$\implies \cos x=\pm1$

This implies that: $x=0,\pi,2\pi$

If $\tan^2x=-2$, x is not defined for all real values.

Therefore the required solution on the given interval $0\le x\le2\pi$ is  $x=0,\pi,2\pi$