If f(θ) = a cos(bθ), then from the first condition we find
f(0) = 3 ⇒ a cos(0) = 3 ⇒ a = 3
Together with the other conditions, it's evident that f(θ) has a period of π, so
2π/b = π ⇒ b = 2
so that
f(θ) = 3 cos(2θ)
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