解三角方程

2012-06-25 2:43 am
1. sin x = sin (π/2 - 2x)
2. (2 sin x - 1)(sin x + 1) = 0

回答 (1)

2012-06-25 5:43 am
✔ 最佳答案
1. Sinx=Sin(π/2-2x)
Sol
Sinx=Sin(π/2-2x)
Sinx=Cos(2x)
Sinx=1-2sin^2 x
2Sin^2 x-Sinx-1=0
(2Sinx-1)(Sinx+1)=0
(1) 2Sinx-1=0
Sinx=1/2
x=nπ+(-1)^n*(π/6),n is a integer
(2) Sinx+1=0
Sinx=-1
x=nπ-(-1)^n*(π/2),n is a integer

2. (2Sinx-1)(Sinx+1)=0
(1) 2Sinx-1=0
Sinx=1/2
x=nπ+(-1)^n*(π/6),n is a integer
(2) Sinx+1=0
Sinx=-1
x=nπ-(-1)^n*(π/2),n is a integer




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