求f(x)=(√3-2cos(2x+π/3))/(1+2sin(2x+π/3))定义域,值域(步骤)
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/04 02:58:21
![求f(x)=(√3-2cos(2x+π/3))/(1+2sin(2x+π/3))定义域,值域(步骤)](/uploads/image/z/6863054-14-4.jpg?t=%E6%B1%82f%28x%29%3D%28%E2%88%9A3-2cos%282x%2B%CF%80%2F3%29%29%2F%281%2B2sin%282x%2B%CF%80%2F3%29%29%E5%AE%9A%E4%B9%89%E5%9F%9F%2C%E5%80%BC%E5%9F%9F%EF%BC%88%E6%AD%A5%E9%AA%A4%EF%BC%89)
求f(x)=(√3-2cos(2x+π/3))/(1+2sin(2x+π/3))定义域,值域(步骤)
求f(x)=(√3-2cos(2x+π/3))/(1+2sin(2x+π/3))定义域,值域(步骤)
求f(x)=(√3-2cos(2x+π/3))/(1+2sin(2x+π/3))定义域,值域(步骤)
1+2sin(2x+π/3)≠0,解得2x+π/3≠2kπ+7π/6 且 2x+π/3≠2kπ+11π/6
x≠kπ-π/4 且 x≠kπ+5π/12
f(x)=(√3-2cos(2x+π/3))/(1+2sin(2x+π/3))
设2x+π/3=θ+π/6
f(x)=(√3-2cos(θ+π/6))/(1+2sin(θ+π/6))
=(√3-√3cosθ+sinθ)/(1+√3sinθ+cosθ)
∵sinθ/(1+cosθ)=(1-cosθ)/sinθ=tan(θ/2)
∴由和比定理得f(x)=tan(θ/2)=tan(x+π/12)
∵x≠kπ-π/4 且 x≠kπ+5π/12
∴f(x)≠-√3/3(问了两次呢…………
f(x)定义域满足:分母1+2sin(2x+π/3)≠0
即:sin(2x+π/3)≠-1/2
则有:2x+π/3≠2kπ-π/6且2x+π/3≠2kπ-5π/6 (k属于Z)
则定义域:{x|x≠kπ-π/4且x≠kπ-7π/12,k属于Z}
设m=sin(2x+π/3),n=cos(2x+π/3)
f(x)等价于:点(1,√3)与点(-2m,2n)...
全部展开
f(x)定义域满足:分母1+2sin(2x+π/3)≠0
即:sin(2x+π/3)≠-1/2
则有:2x+π/3≠2kπ-π/6且2x+π/3≠2kπ-5π/6 (k属于Z)
则定义域:{x|x≠kπ-π/4且x≠kπ-7π/12,k属于Z}
设m=sin(2x+π/3),n=cos(2x+π/3)
f(x)等价于:点(1,√3)与点(-2m,2n)所在直线的斜率
由于:m^2+n^2=1
则:(-2m)^2+(2n)^2=4
故点(-2m,2n),(1,√3)都在圆:x^2+y^2=4上
由图像可知:
在该圆上,定点(1,√3)与动点(-2m,2n)间斜率可为任意值
故f(x)属于R
收起