`sin(x) - 2 = cos(x) - 2` Find all the solutions of the equation in the interval `0,2pi)`.

Textbook Question

Chapter 5, 5.3 - Problem 32 - Precalculus (3rd Edition, Ron Larson).
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gsarora17 | (Level 2) Associate Educator

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`sin(x)-2=cos(x)-2`

`sin(x)-2-cos(x)+2=0` 

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

`(sin(x)-cos(x))/cos(x)=0`

`tan(x)-1=0`

`tan(x)=1`

General solutions for tan(x)=1 are `x=pi/4+pin`

Solutions for the range `0<=x<=2pi` are

`x=pi/4 , (5pi)/4`

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