作业帮函数f(x)=sin2x·sin(π/6)-cos2x·cos(5π/6)在[-π/2,π/2]上的单调递增区间为?
来源：学生作业帮助网 编辑：作业帮 时间：2024/08/02 10:12:19
![函数f(x)=sin2x·sin(π/6)-cos2x·cos(5π/6)在[-π/2,π/2]上的单调递增区间为?](/uploads/image/z/13330608-24-8.jpg?t=%E5%87%BD%E6%95%B0f%28x%29%3Dsin2x%C2%B7sin%28%CF%80%2F6%29-cos2x%C2%B7cos%285%CF%80%2F6%29%E5%9C%A8%5B-%CF%80%2F2%2C%CF%80%2F2%5D%E4%B8%8A%E7%9A%84%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4%E4%B8%BA%3F)
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函数f(x)=sin2x·sin(π/6)-cos2x·cos(5π/6)在[-π/2,π/2]上的单调递增区间为?
函数f(x)=sin2x·sin(π/6)-cos2x·cos(5π/6)在[-π/2,π/2]上的单调递增区间为?
函数f(x)=sin2x·sin(π/6)-cos2x·cos(5π/6)在[-π/2,π/2]上的单调递增区间为?
直接得到：F（x）=sin（2X+π/3）,由于是正弦函数,所以-π/2=