maths (trigonometric)

solve √2 cos^2θ + cosθ = 0
cos θ (√2 cosθ + 1) = 0
cosθ =0 or cosθ =-1/√2
θ =90, 135,225,270
(since cos45=1/√2)

2007-01-31 19:28:25 補充：
雖然閣下明白﹐但我都補充下啦1 任何三角函數的第一象限都是正的(包括cos)2 cos(-θ)=-cosθ 3 cos(180-θ )=-cosθ 4 cos(180+θ )=-cosθ 5 cos(360-θ )=cosθ 所以cos(180-45)=-cos45cos(180+45)=-cos45cos135=cos225=-cos45=-1/√2

其實好似二次方程式咁計
√2 cos²θ + cosθ = 0
設cosθ=k 咁.....
k²√2 + k = 0
抽k
k(k√2 + 1)=0
k=0 or k=-1/√2
∴cosθ=0 or cosθ=-1/√2
θ=90 or θ=135 / 225

√2 cos^2θ + cosθ = 0
(√2 cosθ +1) * (cosθ - 0)

=> √2 cosθ = -1
θ = 135°

but sorry about the second one i couldn't work it out