Question

How do you solve `2tan^2x - 3secx + 3 = 0`?

Answer 1

Call sec x = t. Use the trig identity: 1 + tan^2 x = sec^2 x = t^2

`f(x) = 2(t^2 - 1) - 3t + 3 = 2t^2 - 3t + 1 = 0`
Since (a + b + c ) = 0, one real root is t = 1 and the other is t = c/a = 1/2
Solve the 2 basic trig equations: t = 1 and t = 1/2.

a. t = sec x = 1/cos x = 1 --> cos x = 1 -> x = 2pi

b. t = sec x = 1/cos x = 1/2 -> cos x = 2 (rejected)

Check: x = 2pi -> cos x = 1 -> sec = t = 1.--> tan x = 0

2t^2 - 3t + 1 = 0 -> 2 - 3 + 1 = 0 . OK

2tan^2 - 3sec + 3 = 0 --> 0 - 3 + 3 = 0 OK