题目
Problem
In this question you must show all stages of your working.
Solutions based entirely on calculator technology are not acceptable.
(i) Solve, for −32π<x<32π, the equation
5sin(3x+0.1)+2=0
giving your answers, in radians, to 2 decimal places.
(4)
(ii) Solve, for 0<θ<360∘, the equation
2tanθsinθ=5+cosθ
giving your answers, in degrees, to one decimal place.
(5)
解答
(i)
解法一
思路
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- 化簡方程:
5sin(3x+0.1)+2=0⟹sin(3x+0.1)=−0.4
- 換元與確定範圍:
令 y=3x+0.1。已知 −32π<x<32π,乘以 3 再加 0.1,得到 y 的取值範圍:
−2π+0.1<y<2π+0.1
即 −6.183<y<6.383。
- 求 y 的解:
- 設 siny=−0.4,其主值(principal value)為 α=arcsin(−0.4)≈−0.4115。
- 在一個週期 2π 內,另一解為 π−α=π−(−0.4115)≈3.5531。
- 尋找在範圍 (−2π+0.1,2π+0.1) 內的所有解:
- y1=α≈−0.4115
- y2=π−α≈3.5531
- y3=2π+α≈5.8717
- y4=−π−α≈−2.7301
- 求 x 的解:
利用 x=3y−0.1 計算對應的 x,並保留 2 位小數。
答题过程
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Simplify the equation:
5sin(3x+0.1)+2=sin(3x+0.1)=0−0.4.
Let y=3x+0.1. The range of x is −32π<x<32π, which gives the range of y:
3(−32π)+0.1<−2π+0.1<y<3(32π)+0.1y<2π+0.1.
Now solve siny=−0.4. The principal value of y is:
y=arcsin(−0.4)≈−0.41152 rad.
The other solution in the basic interval [−π,π] is:
y=−π−(−0.41152)=−π+0.41152≈−2.73007 rad.
Adding/subtracting multiples of 2π to find all solutions within the range −2π+0.1<y<2π+0.1:
- For y=−0.41152:
- y=−0.41152
- y=2π−0.41152≈5.87167
- For y=−2.73007:
- y=−2.73007
- y=2π−2.73007 (this is 3.55311 which is within the range)
So the solutions for y are:
y≈−2.73007,−0.41152,3.55311,5.87167.
Now calculate the corresponding values of x using x=3y−0.1:
- For y=−2.73007:
x=3−2.73007−0.1≈−0.94336≈−0.94
- For y=−0.41152:
x=3−0.41152−0.1≈−0.17051≈−0.17
- For y=3.55311:
x=33.55311−0.1≈1.15104≈1.15
- For y=5.87167:
x=35.87167−0.1≈1.92389≈1.92
Therefore, the solutions are:
x=−0.94,−0.17,1.15,1.92(to 2 d.p.).
(ii)
解法一
思路
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- 替換切切:利用關係式 tanθ=cosθsinθ。
2(cosθsinθ)sinθ=5+cosθ⟹cosθ2sin2θ=5+cosθ
- 去分母與統一三角函數:
方程兩邊同乘以 cosθ(需注意 cosθ=0):
2sin2θ=5cosθ+cos2θ
利用 sin2θ=1−cos2θ 將正弦轉換為餘弦:
2(1−cos2θ)=5cosθ+cos2θ⟹3cos2θ+5cosθ−2=0
- 因式分解求解:
將二次三項式因式分解為 (3cosθ−1)(cosθ+2)=0。
由於餘弦值的範圍為 [−1,1],所以 cosθ=−2 無解。
故 cosθ=31。
- 求 θ 的解:
在範圍 0<θ<360∘ 內,餘弦值為正,有兩個解:第一象限角和第四象限角。
答题过程
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Using tanθ=cosθsinθ:
2tanθsinθ=2(cosθsinθ)sinθ=cosθ2sin2θ=5+cosθ5+cosθ5+cosθ.
Multiply by cosθ:
2sin2θ=5cosθ+cos2θ.
Use the identity sin2θ=1−cos2θ:
2(1−cos2θ)=2−2cos2θ=3cos2θ+5cosθ−2=(3cosθ−1)(cosθ+2)=5cosθ+cos2θ5cosθ+cos2θ00.
Since −1≤cosθ≤1, cosθ=−2 has no real solutions.
Thus, we have:
3cosθ−1=cosθ=031.
For 0<θ<360∘:
θ=arccos(31)≈70.5288∘≈70.5∘,
and the other solution is:
θ=360∘−70.5288∘≈289.4712∘≈289.5∘.
Therefore, the solutions are:
θ=70.5∘,289.5∘(to 1 d.p.).