Question

How do you solve `2sec2x - cot2x = tan2x` from 0 to 2pi?

Answer 1

`x=pi/12,x=5*pi/12,x=13*pi/12,x=17*pi/12`

Explanation:

Simplifying and factorizing the given equation we obtain
`(-2+csc(2x))sec(2x)=0`

Answer 2

`pi/12; (13pi)/12, and (5pi)/12 ; (17pi)/12`

Explanation:

Call X = 2x, the equation becomes:
`2/(cos X) - cos X/(sin X) - sin X/(cos X) = 0`
`(2sin X - cos^2 x - sin^2 x)/(sin X.cos x) = 0`
`2sin X - 1 + sin^2 X - sin^2 X = 0`
Condition --> `(sin X.cos X) != 0`
`2sin X = 1` --> `sin X = 1/2` --> `sin 2x = 1/2`
Trig table and unit circle give 2 general solutions for 2x -->
`2x = pi/6 + 2kpi` and `2x = (5pi)/6 + 2kpi`
a. `2x = pi/6 + 2kpi` --> `x = pi/12 + kpi`
b. `2x = (5pi)/6 + 2kpi` --> `x = (5pi)/12 + kpi`
From 0 to `2pi`, there are 4 answers (k = 1):
`pi/12`; and `pi/12 + pi = (13pi)/12`
`(5pi)/12`; and `(5pi)/12 + pi = (17pi)/12`