Question

2sec^2x -3tanx =5 for [0,2pi)
Please solve for the values of x for me.

Answer 1

I am sure bobpursly had already answered this for you, but I could not find the post. I recall he took you as far as the quadratic. Why did you not finish it ??

2sec^2x -3tanx =5 for [0,2pi)

make use of your identities
sec^2 A = 1 + tan^2 A

2(1 + tan^2 x) - 3tanx - 5 = 0
2 + 2tan^2 x - 3tanx - 5 = 0
2tan^2 x - 3tanx - 3 = 0
by the formula:

tanx = (3 ± √33)/4
= appr 2.18614... or -.68614...

using tanx = 2.18614...
x is in quads I or III
x = 1.1418 or x = π + 1.1418 = 4.2834

using tanx = -.68614..
x is in quads II or IV
x = π - .60136... = 2.5102
x - 2π - .60136 = 5.6818 , all 4 answers rounded to 4 decimal at the final stage, not before.

Answer 2

in the line near the end:
x = π - .60136... = 2.5102
it should be

x = π - .60136... = <b>2.5402>/b></b>