a maths 三角函數

解下列各方程
2sin^4x + 7cos^2x = 4
我用以下方法計算
2sin^4x + 7(1-sin^2x) = 4
2sin^4x + 7 - 7sin^2x - 4 = 0
2sin^4x - 7sin^2x + 3 = 0
sin^2x = 3 (reject) or sin^2x = 1/2

咁x應該等如 60, 120, 240, 300degree
但係正確答案係 45, 135, 225, 315degree
我有邊到做錯???
請解答
sol
we must define 0<x<360
sin^2x = 1/2
sinx=sqrt(2)/2 or sinx=-sqrt(2)/2
when sinx=-sqrt(2)/2
x=225degree ,315degree
when sinx=sqrt(2)/2
x=45degree ,135degree
so x=



2009-03-20 22:25:15 補充：
45, 135, 225, 315degree

你步驟無錯~
但係sin^2x = 1/2
sin x=+1/√2 or x=-1/√2
x=45or135 or x=225or315

Your solution for sin^2x = 1/2 is right. However, if this is the case, x should be 45, 135, 225 or 315.

sin 45 = 1/sqrt(1/2) ==> sin^2(45) = 1/2
(sin 135)^2 = 1/2
(sin 225)^2 = 1/2
(sin 315)^2 = 1/2