Verified answer

For all nonzero x in F,

1 = f(1) = f(x x^(-1)) = f(x) f(x^(-1)),

so f(x) cannot be zero. Therefore, f is injective.

replace x = cos ?, so which you get 2 x² - 3 x + a million = 0 fixing that quadratic equation aspects the two recommendations x = a million/2 and x = a million resubstitite and resolve for ? ends up in ? = 0, ? = pi/3 and ? = - pi/3 because of the fact the cosine is periodic with era 2 pi, you could upload multiples of two pi to those answer to get the entire set of recommendations.