已知向量a=(2sin(x+θ/2),根号3),向量b=(cos(x+θ/2),2cos^2(x+θ/2))f(x)=向量a·向量b-根号3,求f(x)的解析式(2)若0<θ<π,求θ使f(x)为偶函数,并求此时f(x)=1,x∈[-π,π]的角的集合
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![已知向量a=(2sin(x+θ/2),根号3),向量b=(cos(x+θ/2),2cos^2(x+θ/2))f(x)=向量a·向量b-根号3,求f(x)的解析式(2)若0<θ<π,求θ使f(x)为偶函数,并求此时f(x)=1,x∈[-π,π]的角的集合](/uploads/image/z/1746954-18-4.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8Fa%3D%EF%BC%882sin%EF%BC%88x%2B%CE%B8%2F2%EF%BC%89%2C%E6%A0%B9%E5%8F%B73%EF%BC%89%2C%E5%90%91%E9%87%8Fb%3D%EF%BC%88cos%EF%BC%88x%2B%CE%B8%2F2%EF%BC%89%2C2cos%5E2%28x%2B%CE%B8%2F2%29%EF%BC%89f%EF%BC%88x%29%3D%E5%90%91%E9%87%8Fa%C2%B7%E5%90%91%E9%87%8Fb-%E6%A0%B9%E5%8F%B73%2C%E6%B1%82f%EF%BC%88x%EF%BC%89%E7%9A%84%E8%A7%A3%E6%9E%90%E5%BC%8F%282%29%E8%8B%A50%EF%BC%9C%CE%B8%EF%BC%9C%CF%80%EF%BC%8C%E6%B1%82%CE%B8%E4%BD%BFf%EF%BC%88x%EF%BC%89%E4%B8%BA%E5%81%B6%E5%87%BD%E6%95%B0%EF%BC%8C%E5%B9%B6%E6%B1%82%E6%AD%A4%E6%97%B6f%EF%BC%88x%EF%BC%89%3D1%EF%BC%8Cx%E2%88%88%5B-%CF%80%EF%BC%8C%CF%80%5D%E7%9A%84%E8%A7%92%E7%9A%84%E9%9B%86%E5%90%88)
已知向量a=(2sin(x+θ/2),根号3),向量b=(cos(x+θ/2),2cos^2(x+θ/2))f(x)=向量a·向量b-根号3,求f(x)的解析式(2)若0<θ<π,求θ使f(x)为偶函数,并求此时f(x)=1,x∈[-π,π]的角的集合
已知向量a=(2sin(x+θ/2),根号3),向量b=(cos(x+θ/2),2cos^2(x+θ/2))
f(x)=向量a·向量b-根号3,求f(x)的解析式
(2)若0<θ<π,求θ使f(x)为偶函数,并求此时f(x)=1,x∈[-π,π]的角的集合
已知向量a=(2sin(x+θ/2),根号3),向量b=(cos(x+θ/2),2cos^2(x+θ/2))f(x)=向量a·向量b-根号3,求f(x)的解析式(2)若0<θ<π,求θ使f(x)为偶函数,并求此时f(x)=1,x∈[-π,π]的角的集合
向量a·向量b=(2sin(x+θ/2),根号3)·(cos(x+θ/2),2cos^2(x+θ/2))
=2sin(x+θ/2)cos(x+θ/2)+根号32cos^2(x+θ/2),
=sin(2x+θ)+根号3(1+cos(2x+θ))
f(x)=向量a·向量b-根号3=sin(2x+θ)+根号3(1+cos(2x+θ)-根号3
f(x)=sin(2x+θ)+根号3cos(2x+θ)
=2sin(2x+pai/3)