已知函数f(x)=sin(2x+π/3),求函数g(x)=log(1/2)f(x)的单调增区间 2kπ+π/2<2x+π/3<2kπ+π的<怎么来的,为什么不能≤;还有2kπ+π中π怎么得的,为什么不是3π/2?
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![已知函数f(x)=sin(2x+π/3),求函数g(x)=log(1/2)f(x)的单调增区间 2kπ+π/2<2x+π/3<2kπ+π的<怎么来的,为什么不能≤;还有2kπ+π中π怎么得的,为什么不是3π/2?](/uploads/image/z/8947426-58-6.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3Dsin%EF%BC%882x%2B%CF%80%2F3%EF%BC%89%2C%E6%B1%82%E5%87%BD%E6%95%B0g%EF%BC%88x%EF%BC%89%3Dlog%EF%BC%881%2F2%EF%BC%89f%EF%BC%88x%EF%BC%89%E7%9A%84%E5%8D%95%E8%B0%83%E5%A2%9E%E5%8C%BA%E9%97%B4+2k%CF%80%2B%CF%80%2F2%EF%BC%9C2x%2B%CF%80%2F3%EF%BC%9C2k%CF%80%2B%CF%80%E7%9A%84%EF%BC%9C%E6%80%8E%E4%B9%88%E6%9D%A5%E7%9A%84%2C%E4%B8%BA%E4%BB%80%E4%B9%88%E4%B8%8D%E8%83%BD%E2%89%A4%EF%BC%9B%E8%BF%98%E6%9C%892k%CF%80%2B%CF%80%E4%B8%AD%CF%80%E6%80%8E%E4%B9%88%E5%BE%97%E7%9A%84%2C%E4%B8%BA%E4%BB%80%E4%B9%88%E4%B8%8D%E6%98%AF3%CF%80%2F2%3F)
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