The number of real solutions of the equation x−sinx=0 is
Let f(x)=x−sinx. then
f′(x)=1 cosx≥0 for all x∈R−{2nπ:n∈Z;
⇒f(x) is strictly increasing on nR−{2nπ:n∈Z}
⇒ f(x)>f(0) for all x>0,x≠2nπ,n∈N
and
f(x)<f(0) for all x<0,x≠−2nπ,n∈N
Thus x>sinx for all x>0, x≠2nπ,n∈N
and x<sinx and for all x<0, x≠−2nπ,n∈N
Thus, x−sinx≠0 for all x∈R−{2nπ:n∈Z}
Hower , x−sinx=0 for x=0 only