Question

How do you solve `2tan^2x+sec^2x=2`?

Answer 1

This can be written as:

`(2sin^2x)/cos^2x + 1/cos^2x = 2`, using the identities `tantheta = sin theta/costheta` and `sectheta = 1/costheta`

Simplifying:

`(2sin^2x+ 1)/cos^2x = 2`

`2sin^2x + 1 = 2cos^2x`

Apply the identity `sin^2x + cos^2x = 1 -> sin^2x = 1 - cos^2x` to the left-hand side.

`2(1 - cos^2x) + 1 = 2cos^2x`

`2 - 2cos^2x + 1 = 2cos^2x`

`3 = 4cos^2x`

`3/4 = cos^2x`

`+-sqrt(3)/2 = cosx`

`x = pi/6, (5pi)/6, (7pi)/6 and (11pi)/6`

Note that these solutions are only those that are in the interval `0 ≤ x < 2pi`.

Hopefully this helps!