cos(2x+π/3)+sin^2x-1/2 求x在 0到π之间时,函数单调递减区间

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y=cos(2x+π/3)+sin²x -1/2
=cos2xcos(π/3)-sin2xsin(π/3)+(1-cos2x)/2 -1/2
=-(√3/2)sin2x
令 -π/2+2kπ≤2x≤π/2+2kπ,
解得 -π/4+kπ≤x≤π/4+kπ,k 是整数
因为 0≤x≤π,从而 k取0,1,得函数单调递减区间为[0,π/4]和[3π/4,π]
======以下答案可供参考======
供参考答案1:
∵cos(2x+π/3)+sin^2x-1/2 =cos2xcosπ/3-sin2xsinπ/3+sin^2x-1/2 =1/2cos2x-(√3/2)sin2x+sin^2x-1/2=1/2-sin^2-(√3/2)sin2x+sin^2x-1/2=-(√3/2)sin2x
∴x在 0到π之间时，函数单调递减区间是[0,π/4]∪[3π/4,π].