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CIE 9709 2025 June Paper 11 Q1

A Level / CIE / P1

CIE 9709 2025 June Paper 11 Paper · Question 1

题目

Problem

Solve the equation 6sin⁡θ=1+2sin⁡θ6\sin\theta = 1 + \frac{2}{\sin\theta} for −180∘<θ<180∘-180^\circ < \theta < 180^\circ.

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题目中文翻译

在 −180∘<θ<180∘-180^\circ < \theta < 180^\circ 范围内,解方程 6sin⁡θ=1+2sin⁡θ6\sin\theta = 1 + \frac{2}{\sin\theta}。

[4]

解答

解法一

思路

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原式要求 sin⁡θ≠0\sin\theta\ne0。乘以 sin⁡θ\sin\theta 后解二次方程,再根据两个正弦值的符号找齐角度。

答题过程

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Multiplication by sin⁡θ\sin\theta gives

6sin⁡2θ−sin⁡θ−2=(2sin⁡θ+1)(3sin⁡θ−2)=0.6\sin^2\theta-\sin\theta-2 =(2\sin\theta+1)(3\sin\theta-2)=0.

For sin⁡θ=−12\sin\theta=-\frac12, the two angles in the interval are −150∘-150^\circ and −30∘-30^\circ. For sin⁡θ=23\sin\theta=\frac23, the principal angle is α=sin⁡−1(2/3)\alpha=\sin^{-1}(2/3) and the other is 180∘−α180^\circ-\alpha. These give 41.8∘41.8^\circ and 138.2∘138.2^\circ. All four have nonzero sine.