Maths唔識抵死題(new2)

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.

the q should be cos 2 y + 3 sin y +1 =0

the ans = 210 degrees and 330 degrees, try again!!