Problem:
How many angles θ with 0≤θ≤2π satisfy log(sin(3θ))+log(cos(2θ))=0 ?
Answer Choices:
A. 0
B. 1
C. 2
D. 3
E. 4
Solution:
Suppose θ satisfies the equation. Then log(sin(3θ)⋅cos(2θ))=0. This implies sin(3θ)⋅cos(2θ)=1, so applying the Product-to-Sum Formula
sina⋅cosb=21(sin(a+b)+sin(a−b))
gives 21(sin(5θ)+sinθ)=1. Then sin(5θ)+sinθ=2, so sin(5θ)=sinθ=1, and the only possible solution is θ=2π.
However, this solution is not valid because sin(3θ) is equal to -1 , and log(sin(3θ)) is not defined.
OR
As in the first solution, sin(3θ)⋅cos(2θ)=1. Then either sin(3θ)=1 and cos(2θ)=1, or sin(3θ)=−1 and cos(2θ)=−1. However, only the first case is possible because otherwise the logarithms in the equation are not defined. In the first case, 3θ∈{2π,25π,29π} and 2θ∈{0,2π,4π}, yielding θ∈{6π,65π,23π} and θ∈{0,π,2π}. The two sets have no common values for θ, so the first case yields no solutions.
The problems on this page are the property of the MAA's American Mathematics Competitions