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 and solutions on this page are the property of the MAA's American Mathematics Competitions