已知f(x)=2sin(x+θ/2)cos(x+θ/2)+2√3cos^2(x+θ/2)-√3问:若0≤θ≤π,且函数f(x)为偶函数;求满足方程f(x)=1,且x∈[0,π]的x的集合.答案是X=5π/6或X=π/6我就是算不出π/6
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/22 09:43:57
![已知f(x)=2sin(x+θ/2)cos(x+θ/2)+2√3cos^2(x+θ/2)-√3问:若0≤θ≤π,且函数f(x)为偶函数;求满足方程f(x)=1,且x∈[0,π]的x的集合.答案是X=5π/6或X=π/6我就是算不出π/6](/uploads/image/z/4018838-14-8.jpg?t=%E5%B7%B2%E7%9F%A5f%EF%BC%88x%EF%BC%89%3D2sin%EF%BC%88x%2B%CE%B8%2F2%EF%BC%89cos%EF%BC%88x%2B%CE%B8%2F2%EF%BC%89%2B2%E2%88%9A3cos%5E2%EF%BC%88x%2B%CE%B8%2F2%EF%BC%89-%E2%88%9A3%E9%97%AE%EF%BC%9A%E8%8B%A50%E2%89%A4%CE%B8%E2%89%A4%CF%80%2C%E4%B8%94%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%E4%B8%BA%E5%81%B6%E5%87%BD%E6%95%B0%3B%E6%B1%82%E6%BB%A1%E8%B6%B3%E6%96%B9%E7%A8%8Bf%EF%BC%88x%EF%BC%89%3D1%2C%E4%B8%94x%E2%88%88%5B0%2C%CF%80%5D%E7%9A%84x%E7%9A%84%E9%9B%86%E5%90%88.%E7%AD%94%E6%A1%88%E6%98%AFX%3D5%CF%80%2F6%E6%88%96X%3D%CF%80%2F6%E6%88%91%E5%B0%B1%E6%98%AF%E7%AE%97%E4%B8%8D%E5%87%BA%CF%80%2F6)
已知f(x)=2sin(x+θ/2)cos(x+θ/2)+2√3cos^2(x+θ/2)-√3问:若0≤θ≤π,且函数f(x)为偶函数;求满足方程f(x)=1,且x∈[0,π]的x的集合.答案是X=5π/6或X=π/6我就是算不出π/6
已知f(x)=2sin(x+θ/2)cos(x+θ/2)+2√3cos^2(x+θ/2)-√3
问:若0≤θ≤π,且函数f(x)为偶函数;求满足方程f(x)=1,且x∈[0,π]的x的集合.
答案是X=5π/6或X=π/6
我就是算不出π/6
已知f(x)=2sin(x+θ/2)cos(x+θ/2)+2√3cos^2(x+θ/2)-√3问:若0≤θ≤π,且函数f(x)为偶函数;求满足方程f(x)=1,且x∈[0,π]的x的集合.答案是X=5π/6或X=π/6我就是算不出π/6
f(x)=2sin(x+θ/2)cos(x+θ/2)+2√3cos^2(x+θ/2)-√3
=sin(2x+θ)+√3(1+cos(2x+θ))-√3
=sin(2x+θ)+√3cos(2x+θ)
=2sin(2x+θ+π/3)
f(x)是偶函数
那么θ+π/3=π/2+kπ(k∈Z)
又0≤θ≤π
那么θ=π/6
所以f(x)=2sin(2x+π/6+π/3)=2cos2x
令f(x)=1得2cos2x=1
那么cos2x=1/2
故2x=2kπ±π/3(k∈Z)
即x=kπ±π/6(k∈Z)
且x∈[0,π]
那么x=π/6或5π/6
如果不懂,请Hi我,祝学习愉快!