2024 AMC 12A 第 8 题

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8.

有多少个满足 0θ2π0\le\theta\le2\pi 的角 θ\theta 使得 log(sin(3θ))+log(cos(2θ))=0\log(\sin(3\theta))+\log(\cos(2\theta))=0

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

答案:A
知识点:对数三角学极限情形界定
难度评级:1480
解答:

方程表示 sin(3θ)cos(2θ)=1\sin(3\theta)\cos(2\theta)=1,且两个因子都必须为正(对数才有定义)。 因为 sin(3θ)1\sin(3\theta)\le1cos(2θ)1\cos(2\theta)\le1,它们的乘积为 11 只能在 sin(3θ)=1\sin(3\theta)=1cos(2θ)=1\cos(2\theta)=1 同时成立时发生。但 cos(2θ)=1\cos(2\theta)=1 强制 θ{0,π,2π}\theta\in\{0,\pi,2\pi\},此时 sin(3θ)=01\sin(3\theta)=0\ne1。没有角满足条件。 因此正确答案是 A

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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