`sec(x) + tan(x) = 1` Find all the solutions of the equation in the interval `0,2pi)`.

Textbook Question

Chapter 5, 5.3 - Problem 38 - Precalculus (3rd Edition, Ron Larson).
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gsarora17 | (Level 2) Associate Educator

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`sec(x)+tan(x)=1`

`1/cos(x)+sin(x)/cos(x)=1`

`1+sin(x)=cos(x)`

`1+sin(x)-cos(x)=0`

`1+cos(pi/2-x)-cos(x)=0`

`1+(2sin((pi/2-x+x)/2)sin((x-pi/2+x)/2)=0`

`1+2sin(pi/4)sin(x-pi/4)=0`

`1+2(1/sqrt(2))(cos(pi/2-x+pi/4)=0`

`1+sqrt(2)(cos((3pi)/4-x)=0`

`1+sqrt(2)(-cos(pi-((3pi)/4-x))=0`

`1-sqrt(2)cos(pi/4+x)=0`

`cos(pi/4+x)=1/sqrt(2)`

General solutions for cos(pi/4+x)=1/ are,

`pi/4+x=pi/4+2pin , (7pi)/4+2pin`

solving above,

`x=2pin , x=(6pi)/4+2pin`

solutions in the range `0<=x<=2pi`  are,

`x=0 ,2pi , (3pi)/2`

since the equation is undefined for x=3pi/2,

Therefore solutions are x=0 and 2pi

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