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solve the equation Sec^2x-Tan x=1 within [0,2pi)

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Sec^2x-Tan x=1 within [0,2pi)
asked Mar 11, 2014 in TRIGONOMETRY by harvy0496 Apprentice

1 Answer

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Given trigonometric equation is Sec ^ 2 x - Tan x = 1

Interval [ 0 , 2 pi ]

Sec ^ 2 x - Tan x = 1 -----------> ( 1 )

We know that Sec ^ 2 x - Tan ^ 2 x = 1

Sec ^ 2 = 1 +Tan ^ 2 x

Substitute the sec ^ 2 x in eq ( 1 )

1 +Tan ^ 2 x - Tan x = 1

1 +Tan ^ 2 x - Tan x - 1 = 0

Tan ^ 2 x - Tan x = 0

( Tan x -  1 )Tan x = 0

( Tan x -  1 ) = 0 or Tan x = 0

Tan x = 0 gives x = 0 , pie

( Tan x -  1 ) = 0

Tan x =1 gives x = ∏ / 4 or  5∏ / 4.

Hence the solution will be x = 0 , ∏  , ∏ / 4 or  5∏ / 4.

answered Apr 1, 2014 by friend Mentor

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