f(x)=sin(2x-5/6π)+2cos2x化简
问题描述：
f(x)=sin(2x-5/6π)+2cos2x化简
并求出f(x)的单调增区间
答
解析：
f(x)=sin(2x-5/6π)+2cos2x
=sin2x*cos(5/6π)- cos2x*sin(5/6π) +2cos2x
=sin2x*[-cos(π/6)] -cos2x*sin(π/6) +2cos2x
=sin2x*(-√3/2) - cos2x*(1/2) +2cos2x
=sin2x*(-√3/2) - cos2x*(3/2)
=-√3*[sin2x*(1/2) +cos2x*(√3/2)]
=-√3*sin（2x+ π/3）
则可知当π/2 + 2kπ≤2x+ π/3≤3π/2 + 2kπ即π/12 + kπ≤2x≤7π/12 + kπ,k属于Z时,正弦型函数y=sin(2x+π/3）是减函数,此时函数f(x)是增函数
所以f(x)的单调增区间是[π/12 + kπ,7π/12 + kπ],k属于Z