设函数F(X)=√2/2cos(2x+π/4)+sin²x,求f(x)的单调区间
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![设函数F(X)=√2/2cos(2x+π/4)+sin²x,求f(x)的单调区间](/uploads/image/z/988993-1-3.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0F%28X%29%3D%E2%88%9A2%2F2cos%282x%2B%CF%80%2F4%29%2Bsin%26%23178%3Bx%2C%E6%B1%82f%28x%29%E7%9A%84%E5%8D%95%E8%B0%83%E5%8C%BA%E9%97%B4)
设函数F(X)=√2/2cos(2x+π/4)+sin²x,求f(x)的单调区间
设函数F(X)=√2/2cos(2x+π/4)+sin²x,求f(x)的单调区间
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