標題:
F4 maths
發問:
solve the following equations for 0度<x<360 度(correct ot 1 decimal place if necessary)1. 〔3-2tan(x+30度)〕．tan(60度-x)=12. sin^2 x-2sinx cosx -cos^2 x=03(a) show that tan^2 x -1-3cos^2 x=(1-2cos^2 x-3cos^4 x)/cos^2 x (b) hence, solve 3cos^2... 顯示更多 solve the following equations for 0度<x<360 度(correct ot 1 decimal place if necessary) 1. 〔3-2tan(x+30度)〕．tan(60度-x)=1 2. sin^2 x-2sinx cosx -cos^2 x=0 3(a) show that tan^2 x -1-3cos^2 x=(1-2cos^2 x-3cos^4 x)/cos^2 x (b) hence, solve 3cos^2 x=tan^2 x-1 for 0度<x<360 度(correct ot 1 decimal place) 更新: 係0度小過x 小過360度
最佳解答:
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(1) [3 - 2 tan (x + 30)] tan (60 - x) = 1 [3 - 2 tan (x + 30)] [1/tan (90 - (60 - x))] = 1 [3 - 2 tan (x + 30)] [1/tan (x + 30)]= 1 3/tan (x + 30) - 2 = 1 3/tan (x + 30) = 3 tan (x + 30) = 1 x + 30 = 45 or 225 x = 15 or 195 (2) sin2 x - 2 sin x cos x - cos2 x = 0 2 sin x cos x = cos2 x - sin2 x sin 2x = cos 2x tan 2x = 1 2x = 45, 225, 405 or 585 x = 22.5, 112.5, 202.5 or 292.5 (3a) Consider (tan2 x - 1 - 3 cos2 x)cos2 x = sin2 x - cos2 x - 3cos4 x = 1 - 2cos2 x - 3cos4 x So tan2 x - 1 - 3 cos2 x = (1 - 2cos2 x - 3cos4 x)/cos2 x (b) 3 cos2 x = tan2 x - 1 tan2 x - 1 - 3 cos2 x = 0 1 - 2cos2 x - 3cos4 x = 0 (1 - 3cos2 x)(1 + cos2 x) = 0 cos2 x = 1/3 or -1 (rejected) cos x = 1/√3 or -1/√3 x = 54.7, 125.3, 234.7 or 305.3 2010-02-20 18:36:37 補充： (2) sin2 x - 2 sin x cos x - cos2 x = 0 -2 sin x cos x = cos2 x - sin2 x -sin 2x = cos 2x tan 2x = -1 2x = 135, 315, 495 or 675 x = 67.5, 167.5, 247.5 or 337.5
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