Question

Question #7d61a

Answer 1

`x=pi/4+(npi)/2` where `ninZZ`

Explanation:

Question: `sec^2x+tan^2x=3tan^2x`

We can move the `tan^2x` to the right-hand side:
`sec^2x=2tan^2x`

Note that `secx=1/cosx` and `tanx=sinx/cosx`
So we have:
`1/cos^2x=(2sin^2x)/cos^2x`
`1/cos^2x-(2sin^2x)/cos^2x=0`
`(1-2sin^2x)/cos^2x=0`

Note the double angle identity for `cos(2x)=cos^2x-sin^2x=1-2sin^2x`
So we have:
`(cos^2x-sin^2x)/cos^2x=0`

`cos^2x/cos^2x-sin^2x/cos^2x=0`

`1-sin^2x/cos^2x=0`

`1-tan^2x=0`

`tan^2x=1`

`tanx=+-sqrt1`

`tanx=+-1`

Therefore,
`x=pi/4+(npi)/2` where `ninZZ`