Question

Question #e4bf6

Answer 1

`x=0, pi/2, pi, (3pi)/2`
`x=(3 pi)/8, (7pi)/8, (11pi)/8, (15pi)/8`

Explanation:

`(sec2x)^2 -1`equals `(tan 2x)^2`. The given equation thus becomes tan 2x (tan 2x +1)=0

It is thus tan 2x=0 or tan 2x= -1

Tan 2x=0 gives `2x=0, pi, 2pi,3pi` that is `x=0, pi/2, pi, (3pi)/2`

Tan 2x=-1, gives `2x= (3pi)/4, (7pi)/4, (11pi)/4, (15pi)/4,` so that

`x=(3 pi)/8, (7pi)/8, (11pi)/8, (15pi)/8`