AMC 10 · 2024 · #8

Grade 11 algebra
logarithm-propertiesdomain-restrictiontrigonometric-ratiosperiodic-functionextreme-principle extreme-principlesystematic-enumeration ↑ Prerequisites: logarithm-propertiestrigonometric-ratios
📏 Medium solution 💡 3 insights
Problem
The equation log(sin(3θ)) + log(cos(2θ)) = 0 is given, and θ is allowed to be any angle from 0 to 2π inclusive. Count how many such angles make the equation true.

Pick an answer.

(A)
0
(B)
1
(C)
2
(D)
3
(E)
4
How to solve
Strategy Extreme Principle

There is no way to peel θ out of log(sin(3θ)) + log(cos(2θ)) = 0 by ordinary algebra, so the plan is to squeeze the equation from both sides until only a short list of candidates is left. Tool #15 (Organize Information in More Ways) does the rewriting: a sum of two logs is the log of a product, and a log that equals 0 means its input is exactly 1, so the log equation becomes the single statement sin(3θ)cos(2θ) = 1. Tool #14 (Extreme Principle) does the real work. Sine and cosine can never exceed 1, and the domain of the logarithm forces both factors to be positive, so a product equal to 1 is only possible when each factor is sitting exactly at its own maximum. That is a boundary case, and boundary cases come with very few candidates. Tool #3 (Eliminate Possibilities) closes it out: solve the easier of the two maximum conditions, write down the handful of angles it permits, and test each one against the other condition.

1STEP 1

Find where the logs are legal

Both quantities must be positive.

sin(3θ) > 0 and cos(2θ) > 0
2STEP 2

Turn the log equation into a product

Merged, their product is 1.

log(sin(3θ)cos(2θ)) = 0 ⟹ sin(3θ)cos(2θ) = 1
3STEP 3

A product of 1 pins both factors

Both must be maximal, so each equals 1.

1 = sc ≤ s ≤ 1 ⟹ sin(3θ) = 1 and cos(2θ) = 1
4STEP 4

List every angle with cosine at its peak

Only three candidates: 0, π, 2π.

2θ ∈ {0, 2π, 4π} ⟹ θ ∈ {0, π, 2π}
5STEP 5

Test the candidates and count

All three fail the sine test, leaving 0 angles.

θ ∈ {0, π, 2π} ⟹ 3θ ∈ {0, 3π, 6π} ⟹ sin(3θ) = 0 ≠ 1 → (A)
Answer
0
The same conclusion should appear when the two conditions are attacked from the other end. sin(3θ) = 1 needs 3θ = π/2 + 2π n, so θ = ((4n+1)π)/6, and inside [0, 2π] that gives θ = π/6, 5π/6, and 3π/2. At the first two, cos(2θ) = 1/2; at the third, cos(2θ) = -1. None of them is 1, which agrees with the count reached above. A numerical sweep says the same thing: scanning θ across [0, 2π] on a grid of 2 × 10⁷ points and keeping only the points where both sin(3θ) > 0 and cos(2θ) > 0, the product sin(3θ)cos(2θ) never rises above about 0.6635, reached near θ ≈ 0.3519. So the sum of the two logs tops out around -0.178 and never reaches 0; no grid point came within 10⁻⁶ of a solution. The near miss worth naming is θ = π/2, where the product really does equal 1 — but as (-1)(-1), with both logarithms undefined. That is the trap the answer choices are built around.
💡Key takeaway

If two positive numbers that are each at most 1 multiply to 1, both have to be exactly 1 — and cos(2θ) = 1 only at multiples of π, which is precisely where sin(3θ) is stuck at 0.

  • Find where the logs are legal
  • Turn the log equation into a product
  • A product of 1 pins both factors
  • List every angle with cosine at its peak
  • Test the candidates and count