Proving Infinite Solutions for sin(x)=b in R using Contradiction Method

[madah12]

Homework Statement

I want to practice proofs by contradiction I am trying to prove that sin(x)=b has infinite distinct solutions in R for every b in [-1,1]

Homework Equations

The Attempt at a Solution

assume it has finite amount of solutions called set Z let k be in real number belong in Z such that sin(k)=b , k >= x for every x in Z that is because if it is a finite set it must have a largest element of Z
there cannot be any x that belongs in Z such that x>k
however sin(k+2pi) = sin(k)*cos(2pi)+sin(2pi)*cos(k) = sin(k)=b
thus k+2pi is a solution therefore belongs inZ and k+2pi>k which means there is an x in the Z such that that x>k which contradicts the assumption that k is the largest element of Z
thus there is no real number k that can fulfill the condition of being the largest element in Z so Z has elements that grow without bound thus Z has infinite solution set

 

[icystrike]

Homework Statement

I want to practice proofs by contradiction I am trying to prove that sin(x)=b has infinite distinct solutions in R for every b in [-1,1]

Homework Equations

The Attempt at a Solution

assume it has finite amount of solutions called set Z let k be in real number belong in Z such that sin(k)=b , k >= x for every x in Z that is because if it is a finite set it must have a largest element of Z
there cannot be any x that belongs in Z such that x>k
however sin(k+2pi) = sin(k)*cos(2pi)+sin(2pi)*cos(k) = sin(k)=b
thus k+2pi is a solution therefore belongs inZ and k+2pi>k which means there is an x in the Z such that that x>k which contradicts the assumption that k is the largest element of Z
thus there is no real number k that can fulfill the condition of being the largest element in Z so Z has elements that grow without bound thus Z has infinite solution set

Your reasoning sounds perfectly good and elegant to me.. shall wait for more comments..

 

[madah12]

any suggestions on how to improve it? any comment?