题目
Problem
Solve the equation
8 sin 4 θ − 6 cos 2 θ + 1 = 0 8\sin^4\theta-6\cos^2\theta+1=0 8 sin 4 θ − 6 cos 2 θ + 1 = 0 for 0 ≤ θ ≤ 2 π 0\le\theta\le2\pi 0 ≤ θ ≤ 2 π
(5)
题目中文翻译
在 0 ≤ θ ≤ 2 π 0\le\theta\le2\pi 0 ≤ θ ≤ 2 π
8 sin 4 θ − 6 cos 2 θ + 1 = 0. 8\sin^4\theta-6\cos^2\theta+1=0. 8 sin 4 θ − 6 cos 2 θ + 1 = 0. (5)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。
解答
解法一
思路
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把余弦平方换成正弦平方后，令 u = sin 2 θ u=\sin^2\theta u = sin 2 θ
答题过程
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Using cos 2 θ = 1 − sin 2 θ \cos^2\theta=1-\sin^2\theta cos 2 θ = 1 − sin 2 θ 8 u 2 + 6 u − 5 = 0 8u^2+6u-5=0 8 u 2 + 6 u − 5 = 0 u = sin 2 θ u=\sin^2\theta u = sin 2 θ
( 4 u + 5 ) ( 2 u − 1 ) = 0. (4u+5)(2u-1)=0. ( 4 u + 5 ) ( 2 u − 1 ) = 0. Only u = 1 / 2 u=1/2 u = 1/2 [ 0 , 1 ] [0,1] [ 0 , 1 ] sin θ = ± 1 / 2 \sin\theta=\pm1/\sqrt2 sin θ = ± 1/ 2
θ = π 4 , 3 π 4 , 5 π 4 , 7 π 4 . \theta=\frac\pi4,\quad\frac{3\pi}4,\quad\frac{5\pi}4,\quad\frac{7\pi}4. θ = 4 π , 4 3 π , 4 5 π , 4 7 π .
解法二
思路
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也可以把正弦的四次方表示成 ( 1 − cos 2 θ ) 2 (1-\cos^2\theta)^2 ( 1 − cos 2 θ ) 2
答题过程
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Put v = cos 2 θ v=\cos^2\theta v = cos 2 θ
8 ( 1 − v ) 2 − 6 v + 1 = 8 v 2 − 22 v + 9 = ( 4 v − 9 ) ( 2 v − 1 ) = 0. \begin{align*}
8(1-v)^2-6v+1=&\,8v^2-22v+9\\
=&\,(4v-9)(2v-1)=0.
\end{align*} 8 ( 1 − v ) 2 − 6 v + 1 = = 8 v 2 − 22 v + 9 ( 4 v − 9 ) ( 2 v − 1 ) = 0. Since 0 ≤ v ≤ 1 0\le v\le1 0 ≤ v ≤ 1 v = 1 / 2 v=1/2 v = 1/2 cos θ = ± 1 / 2 \cos\theta=\pm1/\sqrt2 cos θ = ± 1/ 2 θ = π / 4 , 3 π / 4 , 5 π / 4 , 7 π / 4 \theta=\pi/4,3\pi/4,5\pi/4,7\pi/4 θ = π /4 , 3 π /4 , 5 π /4 , 7 π /4
