Question

If f(x)=sinx+cosx, find the values of n for which (f(x))^n=f(x)

Answer 1

ie when [f(x)]^n - f(x) =0

f(x) [ ( f(x) )^(n-1) -1 ] =0

Answer 2

so f(x) =0 , or [ f(x)]^(n-1) = 1

Answer 3

f(x) =0 --> gives nothing

[f(x)]^(n-1) = 1 --> gives n-1 = 0 , n=1

Answer 4

the answer says 4i, i E N

Answer 5

I got one too but I don't understand how to get 4i