中四AMATH題=.=

2009-02-13 6:17 am
Solve the following equations for 0° <= x <360°

cos(2x + 10°) + cos(2x - 10°)= 0

有冇人識計=.=?

回答 (4)

2009-02-13 7:01 am
✔ 最佳答案
cos(2x + 10°) + cos(2x - 10°)= 0

(cos2x cos10° - sin2x sin10°) + (cos2x cos10° + sin2x sin10°) = 0

cos2x cos10° - sin2x sin10° + cos2x cos10° + sin2x sin10° = 0

2cos2x cos10° = 0

cos2x = 0

2x = 90o, (360° - 90°), (360° + 90°), (720° - 90°)

2x = 90°, 270°, 450°, 630°

x = 45°, 135°, 225°, 315°

2009-02-13 13:53:37 補充:
2cos2x cos10° = 0
2, cos2x, cos10° 三個數相乘等於 0,其中一個數必定會等於 0。
2 不等於 0,而 cos10° 亦不等於 0,所以 cos 2x 一定等於 0

較詳細的寫法是:
2cos2x cos10° = 0
2cos10° ≠ 0,所以 cos2x = 0
2009-02-13 8:07 am
2cos2x cos10° = 0

cos2x = 0 (因為2邊同時左除2cos10°)
2009-02-13 6:31 am
cos(2x+10)+cos(2x-10)=
cos2x cos10-sin2x sin 10 - cos2x cos10-sin2x sin10=0
sin2x sin10=0
sin2x=0
2x = pi n
x=90 n// where n=0,1,2,...

2009-02-13 6:27 am
2x+2x=20°
∵0° ≦x<360°
∴x=≦360°


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