數 學 問 題

Tanθ=(2+Cosθ)/(3Sinθ) (其中0°<=θ<360°)
Sol
Tanθ =(2+Cosθ)/(3Sinθ)
Tanθ*3Sinθ=2+Cosθ
(Sinθ/Cosθ)*3Sinθ=2+Cosθ
3Sin^2 θ=2Cosθ+Cos^2 θ
3－3Cos^2 θ=2Cos+Cos^2θ
4Cos^2 θ+2Cosθ－3=0
Cosθ=[－2+√(4+48)]/8 (負不合)
Cosθ=0.65139
θ=49.35度 or θ=310.65度

(=^o^=)
又係我啦~

根據你打的題目：
tanθ = 2 + cosθ / 3sinθ
tanθ = 2 + (cosθ/cosθ) / (3sinθ/cosθ)
tanθ = 2 + 1/ (3tanθ)
全式乘 3tanθ
3 tan²θ = 6 tanθ + 1
3 tan²θ - 6 tanθ - 1 = 0
3 tan²θ - 6 tanθ + 3 = 4
3 (tan²θ - 2 tanθ + 1) = 4
tan²θ - 2 tanθ + 1 = 4/3
(tanθ - 1)² = 4/3
tanθ - 1 = ±√(4/3)
tanθ = 1±√(4/3)
tanθ = 1± 2/√3
θ = 65.10390936° or 245.1039094° or 171.2060231° or 351.2060231°

如果萬一你打題目的意思是（只是萬一）：
tanθ = (2 + cosθ) / 3sinθ
3sinθ tanθ = (2 + cosθ)
3sinθ (sinθ/cosθ) = (2 + cosθ)
3sin²θ = (2 + cosθ)cosθ
3sin²θ = 2cosθ + cos²θ
3(1-cos²θ) = 2cosθ + cos²θ
3 - 3cos²θ = 2cosθ + cos²θ
4cos²θ + 2cosθ - 3 = 0
cosθ = (-1+√13)/4 or (-1-√13)/4 (rejected)
θ = 49.35368063° or 310.6463194°


2013-10-20 20:47:12 補充：
可以考慮使用解二次方程的公式：

a x² + b x + c = 0
＝＞　x = [ -b ±√(b² - 4ac) ] /(2a)

4cos²θ + 2cosθ - 3 = 0
　x = cosθ
　a = 4
　b = 2
　c = -3

cosθ = [ -2 ±√(2² - 4*4*(-3)) ] /(2*4)
cosθ = [ -2 ±√52 ] /8
cosθ = [ -1 ±√13 ] /4　〔外邊除２，即根號中除４〕