Question

How do you solve `cosxtanx-2cos^2x=-1`?

Answer 1

`pi/6 + 2kpi`
`(5pi)/6 + 2kpi`

Explanation:

`cos x(sin x/(cos x)) = 2cos^2 x - 1`
Simplify by cos x. We get, with condition: cos x diff. to zero, or x diff. to `pi/2 and (3pi)/2`
`sin x = 2cos^2 x - 1 = cos 2x = 1 - 2sin^2 x`
Reminder of trig identity: `cos 2x = 2cos^2 x - 1 = 1 - 2sin^2 x`
Solve this quadratic equation for sin x:
`2sin^2 x + sin x - 1 = 0`
Since a - b + c = 0, use shortcut. The 2 real roots are:
sin x = - 1 and `sin x = - c/a = 1/2`
Use trig table and unit circle -->
a. sin x = - 1 --> arc `x = (3pi)/2`
This solution is rejected because of the above condition.
b. `sin x = 1/2` -->
`x = pi/6` and `x = pi - pi/6 = (5pi)/6`
General answers:
`pi/6 + 2kpi`
`(5pi)/6 + 2kpi`