trigonometry math

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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 -...

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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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