Question

2sec^2x - 3tanx - 5 = 0
over the interval 0 to 2pi

x= 1.14, 4.28, 5.68, and 2.54 right?

Answer 1

2sec^2x - 3tanx - 5 = 0
=> 2(1+tan^2(2x)) - 3tanx - 5 = 0
=> 2tan^2(2x) - 3tanx - 3 = 0
=> (2tan2x + 1)(tanx - 2) = 0
=> 2tan2x = -1, tanx = 2

=> tan2x = -1/2, tanx = 2
=> x = 1.107, 4.249, x = 1.339, 2.91, 4.481, 6.051

Answer 2

2sec^2 x - 3tanx - 5 = 0
2(1+tan^2 x) - 3tanx - 5 = 0
2 + 2tan^2 x - 3tanx - 5 = 0
2tan^2 x - 3tanx - 3 = 0
tanx = (3 ± √33)/4
tanx = 2.18614... or tanx = -.68614...

x = 1.14178... or x = π + 1.14178 = 4.283376
or
x = π - .60136 = 2.5402
x = 2π - .60136 = 5.68182

x = 1.14178 , 2.5402, 4.283376, 5.68182