f is a real-valued function not identically zero, satisfying f(x+y)+f(x−y)=2f(x)⋅f(y)∀x,y∈R⋅f(x)  is definitely

Putting x=0,y=0 in the functional equation we get 2f(0)=2f2(0)⇒f(0)=0,1. But if f(0) is equal to zero then f(x)=0∀x∈R. Thus, f(0)=1. Now putting x=0 in  the functional equation we get, f(y)+f(−y)=2f(0)⋅f(y)⇒f(y)=f(−y). Thus, f(x) is even.