Question

2 tan^2 x-sec x+1 =0​

Answer 1

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

Answer 2

Answer:

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.