作业帮已知函数f(x)=1+sin2x,g(x)=(√2)sin(x+π/4),x∈[-π/2,π/2]（1）求满足f(x)=g(x)的x值的集合（2）求函数f(x)/g(x)的单调递增区间
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![已知函数f(x)=1+sin2x,g(x)=(√2)sin(x+π/4),x∈[-π/2,π/2]（1）求满足f(x)=g(x)的x值的集合（2）求函数f(x)/g(x)的单调递增区间](/uploads/image/z/12555020-20-0.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D1%2Bsin2x%2Cg%28x%29%3D%28%E2%88%9A2%29sin%28x%2B%CF%80%2F4%29%2Cx%E2%88%88%5B-%CF%80%2F2%2C%CF%80%2F2%5D%EF%BC%881%EF%BC%89%E6%B1%82%E6%BB%A1%E8%B6%B3f%28x%29%3Dg%28x%29%E7%9A%84x%E5%80%BC%E7%9A%84%E9%9B%86%E5%90%88%EF%BC%882%EF%BC%89%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%2Fg%28x%29%E7%9A%84%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4)
已知函数f(x)=1+sin2x,g(x)=(√2)sin(x+π/4),x∈[-π/2,π/2]（1）求满足f(x)=g(x)的x值的集合（2）求函数f(x)/g(x)的单调递增区间
已知函数f(x)=1+sin2x,g(x)=(√2)sin(x+π/4),x∈[-π/2,π/2]
（1）求满足f(x)=g(x)的x值的集合
（2）求函数f(x)/g(x)的单调递增区间
已知函数f(x)=1+sin2x,g(x)=(√2)sin(x+π/4),x∈[-π/2,π/2]（1）求满足f(x)=g(x)的x值的集合（2）求函数f(x)/g(x)的单调递增区间
(1)f(x)=1+sin2x=1+2sinxcosx g(x)=(√2)sin(x+π/4)=sinx+cosx
1+2sinxcosx=sinx+cosx
(sinx+cosx)^2=sinx+cosx
sinx+cosx=1或sinx+cox=0
(√2)sin(x+π/4)=0或1
因为x∈[-π/2,π/2]
所以x=-π/4或0或π/2
(2)f(x)/g(x)=[(sinx+cosx)^2]/sinx+cosx=sinx+cosx=(√2)sin(x+π/4)
令-π/2+2kπ≤x+π/4≤π/2+2kπ
-3π/4+2kπ≤x≤π/4+2kπ,k∈Z
因为x∈[-π/2,π/2]
所以单调递增区间为[-π/2,π/4]
楼上的 你没注意x∈[-π/2,π/2]阿
1、
f(x)=g(x)
1+sin2x=(√2)sin(x+π/4),
sin²x+cos²x+2sinxcosx=√2(sinxcosπ/4+cosxsinπ/4)
(sinx+cosx)²=√2(sinx*√2/2+cosx*√2/2)=sinx+cosx
所以(sinx+cosx)²-(sinx+cosx)=0...
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1、
f(x)=g(x)
1+sin2x=(√2)sin(x+π/4),
sin²x+cos²x+2sinxcosx=√2(sinxcosπ/4+cosxsinπ/4)
(sinx+cosx)²=√2(sinx*√2/2+cosx*√2/2)=sinx+cosx
所以(sinx+cosx)²-(sinx+cosx)=0
所以sinx+cosx=0或sinx+cosx=1
sinx+cosx=√2sin(x+π/4)
若√2sin(x+π/4)=0，sin(x+π/4)=0,x+π/4=kπ,x=kπ-π/4
若√2sin(x+π/4)=1，sin(x+π/4)=√2/2,x+π/4=2kπ+π/4或=2kπ+3π/4
则x=2kπ或x=2kπ+π/2
综上
x属于{x|x=kπ-π/4,x=2kπ,x=2kπ+π/2,k是整数}
2、
f(x)=1-cos(π/2+2x)
=1-cos[2(π/4+x)]
=1-[1-2sin²(π/4+x)]
=2sin²(π/4+x)
所以f(x)/g(x)=(√2)sin(x+π/4)
sinx的增区间是(2kπ-π/2,2kπ+π/2)
所以2kπ-π/2
收起
d
f(x)=(sinx+cosx)^2,g(x)=sinx+cosx.看出了这点，问题就很简单了，自己会算吧？