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IAL 2022 May Q5

A Level / Edexcel / P2

IAL 2022 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.

Solve, for 180<θ180-180^\circ < \theta \leqslant 180^\circ, the equation

3tan(θ+43)=2cos(θ+43)3 \tan(\theta + 43^\circ) = 2 \cos(\theta + 43^\circ)

(6)
题目中文翻译

在本题中,你必须展示所有的计算步骤。纯依赖计算器计算的解法是不可接受的。

在区间 180<θ180-180^\circ < \theta \leqslant 180^\circ 内,解方程: 3tan(θ+43)=2cos(θ+43)3 \tan(\theta + 43^\circ) = 2 \cos(\theta + 43^\circ)

解答

解法一

思路

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为方便解题,我们可以引入一个中间变量代换:令 x=θ+43x = \theta + 43^\circ。 原方程化为: 3tanx=2cosx3 \tan x = 2 \cos x

  1. 将正切化为正弦与余弦: 利用商数恒等式(Trigonometric Identity) tanx=sinxcosx\tan x = \frac{\sin x}{\cos x}3sinxcosx=2cosx    3sinx=2cos2x(当 cosx0)3 \frac{\sin x}{\cos x} = 2 \cos x \implies 3 \sin x = 2 \cos^2 x \quad (\text{当 } \cos x \neq 0)

  2. 化为一元二次方程: 利用平方关系恒等式 cos2x=1sin2x\cos^2 x = 1 - \sin^2 x,将方程统一用正弦 sinx\sin x 表达: 3sinx=2(1sin2x)    2sin2x+3sinx2=03 \sin x = 2 (1 - \sin^2 x) \implies 2 \sin^2 x + 3 \sin x - 2 = 0

  3. 因式分解求根: 将上述一元二次方程因式分解: (2sinx1)(sinx+2)=0(2 \sin x - 1)(\sin x + 2) = 0

    • 因为对于任意实数 xx,正弦值的范围是 1sinx1-1 \leqslant \sin x \leqslant 1,因此 sinx+20\sin x + 2 \neq 0(无解)。
    • 于是我们得到唯一合理的解:sinx=12\sin x = \frac{1}{2}
  4. 变换区间并求解: 原自变量 θ\theta 的范围是 180<θ180-180^\circ < \theta \leqslant 180^\circ。 由于 x=θ+43x = \theta + 43^\circ,因此对应 xx 的取值范围为: 180+43<x180+43    137<x223-180^\circ + 43^\circ < x \leqslant 180^\circ + 43^\circ \implies -137^\circ < x \leqslant 223^\circ 在区间 (137,223](-137^\circ, 223^\circ] 内,寻找满足 sinx=12\sin x = \frac{1}{2} 的所有 xx 的解: x=30x=150x = 30^\circ \quad \text{或} \quad x = 150^\circ

  5. 求出最终自变量 θ\theta: 由 θ=x43\theta = x - 43^\circ 还原算出最终答案。

答题过程

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Let x=θ+43x = \theta + 43^\circ. The equation becomes:

3tanx=2cosx3 \tan x = 2 \cos x

Using the identity tanx=sinxcosx\tan x = \frac{\sin x}{\cos x}:

3(sinxcosx)=2cosx3 \left( \frac{\sin x}{\cos x} \right) = 2 \cos x

Multiply both sides by cosx\cos x (noting that cosx0\cos x \neq 0):

3sinx=2cos2x3 \sin x = 2 \cos^2 x

Substitute cos2x=1sin2x\cos^2 x = 1 - \sin^2 x into the equation:

3sinx=2(1sin2x)3 \sin x = 2 \left( 1 - \sin^2 x \right)

3sinx=22sin2x3 \sin x = 2 - 2 \sin^2 x

Rearrange to form a quadratic equation in sinx\sin x:

2sin2x+3sinx2=02 \sin^2 x + 3 \sin x - 2 = 0

Factorize the quadratic equation:

(2sinx1)(sinx+2)=0(2 \sin x - 1)(\sin x + 2) = 0

Since sinx[1,1]\sin x \in [-1, 1], the factor sinx+2=0\sin x + 2 = 0 has no real solutions.

Thus:

2sinx1=0    sinx=122 \sin x - 1 = 0 \implies \sin x = \frac{1}{2}

Now determine the range for xx based on 180<θ180-180^\circ < \theta \leqslant 180^\circ:

180+43<θ+43180+43137<x223\begin{align*} -180^\circ + 43^\circ <&\,\, \theta + 43^\circ \leqslant 180^\circ + 43^\circ \\[2mm] -137^\circ <&\,\, x \leqslant 223^\circ \end{align*}

Solve sinx=12\sin x = \frac{1}{2} for xx in the interval (137,223](-137^\circ, 223^\circ]:

x=30orx=150x = 30^\circ \quad \text{or} \quad x = 150^\circ

Substitute back θ=x43\theta = x - 43^\circ to find the values of θ\theta:

  • For x=30x = 30^\circ: θ=3043=13\theta = 30^\circ - 43^\circ = -13^\circ

  • For x=150x = 150^\circ: θ=15043=107\theta = 150^\circ - 43^\circ = 107^\circ

Since both values fall inside the interval 180<θ180-180^\circ < \theta \leqslant 180^\circ, the required solutions are:

θ=13andθ=107\theta = -13^\circ \quad \text{and} \quad \theta = 107^\circ