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

Textbook Question

Chapter 5, 5.3 - Problem 37 - Precalculus (3rd Edition, Ron Larson).
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gsarora17 | (Level 2) Associate Educator

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`csc(x)+cot(x)=1`

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

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

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

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

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

`1+2sin(pi/4)sin(pi/4-x)=0` 

`1+2(1/sqrt(2))cos(pi/2-(pi/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/`sqrt(2)` are

`pi/4+x=(3pi)/4+2pin , pi/4+x=(5pi)/4+2pin`

solving above,

`x=((3pi)/4-pi/4+2pin) , x=((5pi)/4-pi/4+2pin)`

`x=pi/2+2pin , pi+2pin`

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

`x=pi/2 , pi`

since the equation is undefined for pi

Solution is `pi/2`

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