E. sin(x)=0 when x=0,π, or 2π in the interval 0≤x≤2π. Hence 2cos(2x)+2=0,π, or 2π. Rearranging gives cos(2x)=−1,2π−2, or π−1. cos(2x) achieves the first of these values twice for two different values of x, achieves the second of these values four times, and never achieves the value of π−1 and so the answer is (d).