標題:
兩題三角方程
2(cosx)^2 = sinxcosx (x=63.4,90,243.4,270) 2sinx = tanx (x=0,60,180,300,360) 以上全部0<=x<=360 我需要過程-.-"
最佳解答:
1. 2(cosx)^2 = sinxcosx (cosx)(2cosx - sinx)=0 cosx = 0 or 2cosx = sinx cosx = 0 or tanx = 2 x = 90, 270 or x = 63.4, 243.4 (because tanx is positive) 2. 2sinx = tanx 2sinx = sinx / cosx 2sinxcosx = sinx sinx (2cosx - 1) = 0 sinx = 0 or cosx = 1 / 2 x = 0, 180, 360 or x = 60, 300 (because cosx is positive)
其他解答:7638E7481407D16B
兩題三角方程
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發問:2(cosx)^2 = sinxcosx (x=63.4,90,243.4,270) 2sinx = tanx (x=0,60,180,300,360) 以上全部0<=x<=360 我需要過程-.-"
最佳解答:
1. 2(cosx)^2 = sinxcosx (cosx)(2cosx - sinx)=0 cosx = 0 or 2cosx = sinx cosx = 0 or tanx = 2 x = 90, 270 or x = 63.4, 243.4 (because tanx is positive) 2. 2sinx = tanx 2sinx = sinx / cosx 2sinxcosx = sinx sinx (2cosx - 1) = 0 sinx = 0 or cosx = 1 / 2 x = 0, 180, 360 or x = 60, 300 (because cosx is positive)
其他解答:7638E7481407D16B
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