2024 AMC 12A Problem 8

Below is the professionally curated solution for Problem 8 of the 2024 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 12A solutions, or check the answer key.

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Concepts:logarithmtrigonometrybounding to limit cases

Difficulty rating: 1480

8.

How many angles θ\theta with 0θ2π0\le\theta\le2\pi satisfy log(sin(3θ))+log(cos(2θ))=0?\log(\sin(3\theta))+\log(\cos(2\theta))=0?

00

11

22

33

44

Solution:

The equation means sin(3θ)cos(2θ)=1\sin(3\theta)\cos(2\theta)=1 with both factors positive (for the logs to be defined). Since sin(3θ)1\sin(3\theta)\le1 and cos(2θ)1,\cos(2\theta)\le1, their product is 11 only if sin(3θ)=1\sin(3\theta)=1 and cos(2θ)=1\cos(2\theta)=1 simultaneously. But cos(2θ)=1\cos(2\theta)=1 forces θ{0,π,2π},\theta\in\{0,\pi,2\pi\}, where sin(3θ)=01.\sin(3\theta)=0\ne1. No angle works. Thus, the correct answer is A.

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