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CIE 9709 2025 June Paper 13 Q5

A Level / CIE / P1

CIE 9709 2025 June Paper 13 Paper · Question 5

题目

Problem

Solve the equation 4sin⁡θtan⁡θ=1+5cos⁡θ4\sin\theta\tan\theta = 1 + 5\cos\theta for −180∘<θ<180∘-180^\circ < \theta < 180^\circ.

[6]
题目中文翻译

在 −180∘<θ<180∘-180^\circ < \theta < 180^\circ 范围内,解方程 4sin⁡θtan⁡θ=1+5cos⁡θ4\sin\theta\tan\theta = 1 + 5\cos\theta。

[6]

解答

解法一

思路

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把 tan⁡θ\tan\theta 改写为 sin⁡θ/cos⁡θ\sin\theta/\cos\theta。乘以 cos⁡θ\cos\theta 后,用恒等式把方程化为关于余弦的二次式。

答题过程

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The equation is defined only when cos⁡θ≠0\cos\theta\ne0. Multiplying through by cos⁡θ\cos\theta gives

4sin⁡2θ=cos⁡θ+5cos⁡2θ.4\sin^2\theta=\cos\theta+5\cos^2\theta.

Using sin⁡2θ=1−cos⁡2θ\sin^2\theta=1-\cos^2\theta, we obtain 9cos⁡2θ+cos⁡θ−4=09\cos^2\theta+\cos\theta-4=0. Thus

cos⁡θ=−1+14518orcos⁡θ=−1−14518.\cos\theta=\frac{-1+\sqrt{145}}{18} \quad\text{or}\quad \cos\theta=\frac{-1-\sqrt{145}}{18}.

Both values are in (−1,1)(-1,1) and are nonzero. Cosine is even, so each principal angle and its negative lie in the required interval. The four solutions are −136.4∘-136.4^\circ, −52.2∘-52.2^\circ, 52.2∘52.2^\circ and 136.4∘136.4^\circ.