find the angle x if tan^2 x-4sec x+5=0?
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We have to find x given that : tan^2 x - 4sec x + 5 = 0
(tan x)^2 - 4sec x + 5 = 0
=> ( 1 - (cos x)^2)/( cos x)^2) - 4/ cos x + 5 = 0
let cos x = y
=> (1 - y^2)/y^2 - 4/y + 5 = 0
=> (1 - y^2) - 4y + 5y^2 = 0
=> 4y^2 - 4y - 1 = 0
=> (2y - 1)^2 = 0
=> 2y = 1
=> y = 1/2
=> cos x = 1/2
=> x = arc cos (1/2)
x = pi/3 + 2*pi
The angle x = pi/3 + 2*pi
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We'll substitute (tan x)^2 = (sec x)^2 - 1
We'll re-write the equation:
(sec x)^2 - 1 - 4sec x + 5 = 0
We'll combine like terms:
(sec x)^2 - 4sec x + 4 = 0
We notice that the expression is a perfect square:
(sec x - 2)^2 = 0
We'll put sec x - 2 = 0
sec x = 2
But sec x = 1/cos x => 1/cos x = 2
cos x = 1/2
x = +/- arccos (1/2) + 2kpi
x = +/- (pi/3) + 2kpi
The angles x that verify the equation are {+/- (pi/3) + 2kpi}.
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