设函数F(X)=√2/2cos(2x+π/4)+sin²x,求f(x)的单调区间

先化简cos(2x+π/4),sin^2x →cos(2x+π/4)=√2/2*cos2x-√2/2*sin2x sin^2x=-cos2x/2+1/2 →所以F(x)=cos2x/2-sin2x/2-cos2x/2+1/2=-sin2x/2+1/2 →所以:-π/2+2kπ≤2x≤π/2+2kπ 所以,单减区间:[-π/4+kπ,π/4+kπ],k∈Z →π/2+2kπ≤2x≤3π/2+2kπ 所以单增区间,[π/4+kπ,3π/4+kπ],k∈Z