飛航百科

標題:

Maths唔識抵死題(new2)

發問:

8e) Solve the equations for 0 < x < 360 cos 2 y + 3 sin y +1 =0 show workings 更新: the q should be cos 2 y + 3 sin y +1 =0 the ans = 210 degrees and 330 degrees, try again!!

最佳解答:

is cos 2 y means (cos y)^2 ?? Here cos^2 y means (cos y )^2 Cos ^2 y + 3sin y +1 =0 1 - sin^2 y + 3sin y +1 =0 sin^2 y - 3sin y -2 = 0 ( sin y - 1 )( sin y -2 )=0 sin y =1 or siny = 2 ( reject, since -1 <= sin y < = 1 ) y = 90 deg 2010-09-16 22:51:39 補充： It's something wrong Cos ^2 y + 3sin y +1 =0 1 - sin^2 y + 3sin y +1 =0 sin^2 y - 3sin y -2 = 0 sin y = 3± √ [ (-3)^2-4(-2)] / 2 sin y = 3± √17 / 2 sin y = 3+ √17 / 2 (reject, since -1 ≤ sin y ≤ 1 ) or sin y = 3 - √17 / 2 y = 214.16 / 325.84 deg. 2010-09-16 23:52:03 補充： sorry that I misunderstand the question.. By using cos 2x = cos^2 x - sin^2 x = 1 - 2sin^2 x = 2cos^2 x - 1 cos 2y + 3sin y +1 =0 1 - 2sin^2 y + 3sin y + 1 = 0 2 sin^2 y - 3sin y -2 = 0 (2 sin y+1)(sin y -2) = 0 sin y = -0.5 or sin y = 2 (reject, -1 ≤ sin y ≤ 1 ) y = 210 or 330 deg.

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其他解答:

the q should be cos 2 y + 3 sin y +1 =0 the ans = 210 degrees and 330 degrees, try again!!