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IAL 2023 May Q5

A Level / Edexcel / P3

IAL 2023 May Paper · Question 5

题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

(i) Solve, for 0<x<π0<x<\pi

(x−2)(3sec⁡x+2)=0(x-2)(\sqrt3\sec x+2)=0
(3)

(ii) Solve, for 0<θ<360∘0<\theta<360^\circ

10sin⁡θ=3cos⁡2θ10\sin\theta=3\cos 2\theta
(4)
题目中文翻译

本题中你必须写出解题过程的所有步骤。

不接受完全依赖计算器技术的解法。

(i) 在 0<x<π0<x<\pi 内解方程

(x−2)(3sec⁡x+2)=0(x-2)(\sqrt3\sec x+2)=0

(ii) 在 0<θ<360∘0<\theta<360^\circ 内解方程

10sin⁡θ=3cos⁡2θ10\sin\theta=3\cos 2\theta

解答

(i)

解法一

思路

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乘积为 0，所以分别令两个因子为 0。第一个因子直接给出 x=2x=2；第二个因子转化为 cos⁡x=−32\cos x=-\frac{\sqrt3}{2}，再在 0<x<π0<x<\pi 内选角。

答题过程

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We have

(x−2)(3sec⁡x+2)=0.(x-2)(\sqrt3\sec x+2)=0.

So

x−2=0or3sec⁡x+2=0.x-2=0 \qquad\text{or}\qquad \sqrt3\sec x+2=0.

The first equation gives

x=2.x=2.

For the second equation,

3sec⁡x=−2.\sqrt3\sec x=-2.

So

sec⁡x=−23.\sec x=-\frac2{\sqrt3}.

Hence

cos⁡x=−32.\cos x=-\frac{\sqrt3}{2}.

For 0<x<π0<x<\pi, this gives

x=5π6.x=\frac{5\pi}{6}.

Therefore

x=2orx=5π6.\boxed{x=2} \qquad\text{or}\qquad \boxed{x=\frac{5\pi}{6}}.

(ii)

解法一

思路

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把 cos⁡2θ\cos2\theta 写成 1−2sin⁡2θ1-2\sin^2\theta，这样方程只含 sin⁡θ\sin\theta。解出 sin⁡θ\sin\theta 后，检查哪些值可行，再根据 0<θ<360∘0<\theta<360^\circ 找出所有角。

答题过程

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Start with

10sin⁡θ=3cos⁡2θ.10\sin\theta=3\cos2\theta.

Use

cos⁡2θ=1−2sin⁡2θ.\cos2\theta=1-2\sin^2\theta.

Then

10sin⁡θ=3(1−2sin⁡2θ).10\sin\theta=3(1-2\sin^2\theta).

Expand and rearrange:

10sin⁡θ= 3−6sin⁡2θ,6sin⁡2θ+10sin⁡θ−3= 0.\begin{align*} 10\sin\theta=&\,3-6\sin^2\theta, \\[2mm] 6\sin^2\theta+10\sin\theta-3=&\,0. \end{align*}

Using the quadratic formula,

sin⁡θ= −10±102−4(6)(−3)2⋅6= −10±17212= −5±436.\begin{align*} \sin\theta =&\,\frac{-10\pm\sqrt{10^2-4(6)(-3)}}{2\cdot6} \\[2mm] =&\,\frac{-10\pm\sqrt{172}}{12} \\[2mm] =&\,\frac{-5\pm\sqrt{43}}6. \end{align*}

Now

−5−436<−1,\frac{-5-\sqrt{43}}6<-1,

so it is not possible for sin⁡θ\sin\theta.

Thus

sin⁡θ=−5+436=0.2595⋯ .\sin\theta=\frac{-5+\sqrt{43}}6 =0.2595\cdots.

The reference angle is

sin⁡−1(0.2595⋯ )=15.04⋯∘.\sin^{-1}(0.2595\cdots)=15.04\cdots^\circ.

Since sin⁡θ>0\sin\theta>0 and 0<θ<360∘0<\theta<360^\circ, the solutions are in quadrants I and II:

θ=15.04⋯∘orθ=180∘−15.04⋯∘.\theta=15.04\cdots^\circ \qquad\text{or}\qquad \theta=180^\circ-15.04\cdots^\circ.

Therefore

θ=15.0∘orθ=165∘.\boxed{\theta=15.0^\circ} \qquad\text{or}\qquad \boxed{\theta=165^\circ}.