Question

How do you solve `tan^2x=1-secx` for `0<=x<=2pi`?

Answer 1

`0, (2pi)/3, (4pi)/3, 2pi`

Explanation:

`tan^2 x + sec x - 1 = 0`
`sin^2 x/(cos^2 x) + 1/(cos x) - 1 = 0`
`(sin^2 x + cos x - cos^2 x)/(cos^2 x)` = 0
`sin^2 x - cos^2 x + cos x = 0`. (condition: cos x different to 0)
Replace `sin^2 x` by `(1 - cos^2 x)`, we get
`(1 - cos^2 x) - cos^2 x + cos x = 0`
`- 2cos^2 x + cos x + 1 = 0`
Solve this quadratic equation for cos x.
Since a + b + c = 0, use shortcut. The 2 real roots are:
cos x = 1 and `cos x = c/a = - 1/2`
a. When cos x = 1 --> x = 0 and `x = 2pi`
b. When `cos x = - 1/2` --> `x = +- (2pi)/3`
arc `(4pi)/3` is co-terminal to arc `(-2pi)/3`.
Answers for `(0, 2pi)`
`0, (2pi)/3, (4pi)/3, 2pi`