2 tan^2 x-sec x+1 = 0

Write it with only cos function:

2 * (sin x / cos x) ² - 1/cos x + 1 = 0

(remember that cos x ≠0)

2sin²x/cos²x - 1/cos x + 1 = 0 / *cos²x

2sin²x - cos x + cos² x = 0

2 (1-cos²x) - cos x+cos²x=0

2-2cos²x-cos x + cos²x=0

-cos²x-cos x+2=0 / * (-1)

cos²x+cos x-2=0

I used only these formulas:

tan x = sin x / cos x

sec x = 1/cos x

sin²x + cos² x = 1 (then cos²x = 1-sin²x).

Okay. We've got easy quadratic equation. Substitute t = cos x, remember that - 1≤ t ≤1 and t≠0 is a domain.

t²+t-2=0

∆=1²-4 * (-2) * 1=1+8=9

√∆=3

t1 = (-1+3) / 2 = 1 - it is in domain

t2 = (-1-3) / 2 = - 2 - it's not in domain.

So we reached only cos x = 1.

The solution of this equation is x = 2πk where k is an integral

x = 0, 360 (0 = < x < = 360).

Step-by-step explanation:

2 tan^2 x-sec x+1 = 0

Substituting tan^2 x = sec^2 x - 1:

2 (sec^2 x - 1) - sec x + 1 = 0

2sec^2 x - 2 - sec x + 1 = 0

2 sec^2 x - sec x - 1 = 0

(2 sec x + 1) (sec x - 1) = 0

sec x = - 1/2, sec x = 1

1/cos x = - 1/2, cos x = 1

cos x = - 2 (undefined) = so we ignore this), cos x = 1.

So x = arcos (1) = 0, 360.