標題:
中四AMATH題=.=
發問:
Solve the following equations for 0°
最佳解答:
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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
其他解答:
2cos2x cos10° = 0 cos2x = 0 (因為2邊同時左除2cos10°)|||||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,...|||||2x+2x=20° ∵0° ≦x<360° ∴x=≦360°63D0B758E2D502CC
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