Respuesta :
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:
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.
