find the angle x if tan^2 x-4sec x+5=0?

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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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