標題:
TrigonometryQ2
發問:
Slove the equations for 0°≦x≦360°andgive your ans corr.to 0.1°if necessary.) 1.2cosx-3sinx=3 (cosx+sinx) 2.2cos^2 x=sinxcosx 更新: slove the equations within 0°to 360°
最佳解答:
1) 2cosx ﹣3sinx = 3(cosx + sinx) 2cosx ﹣3sinx = 3cosx + 3sinx 6sinx = - cosx sinx / cosx = -1 / 6 tanx = -1 / 6 x = 170.5 or 350.5 2) 2cos2 x = sinxcosx (2cos2 x)2 = sin2xcos2x 4cos4 x = (1﹣cos2x)cos2x 4cos4 x = cos2x ﹣cos4x 5cos4 x ﹣cos2x = 0 cos2x(5cos2x ﹣1) = 0 cos2x = 0 or 5cos2x ﹣1 = 0 cosx = 0 or cos2x = 1 / 5 x = 90,270 or cosx = √(1 / 5) or -√(1 / 5) x = 90,270 or x = 63.4,116.7,243.4,296.6 ∴x = 63.4,90,116.7,243.4,270,296.6 2008-04-20 21:32:52 補充： 唔好意思，第2題不是這麼複雜，應該是 2cos^2 x = sinxcosx (cosx)(2cosx - sinx)=0 cosx = 0 or 2cosx = sinx cosx = 0 or tanx = 2 x = 90, 270 or x = 63.4, 243.4 x = 63.4, 90,243.4,270
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