y=cos(-2x+π/3)在[0,π])单调递减区间?速求
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y=cos(-2x+π/3)在[0,π])单调递减区间?速求
y=cos(-2x+π/3)在[0,π])单调递减区间?速求
y=cos(-2x+π/3)在[0,π])单调递减区间?速求
【导入】此类题目应该用整体法.
令-2x+π/3=T ∵x∈[0,π]
∴T∈[-5π/3,π/3]
∴y=cosT T∈[-5π/3,π/3]
根据y=cosx的函数性质
y=cosT 在T∈[-5π/3,-π]∪[0,π/3]是递减的
然后 解出关于T=-2x+π/3在T∈[-5π/3,-π]∪[0,π/3]的x的解集
解得x∈[2π/3,π]∪[0,π/6]
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y=cos(-2x+π/3)=cos(2x-π/3)
函数在(2x-π/3)∈[-π+2kπ,2kπ],k∈Z (将2x-π/3看作一个整体)
即x∈[-π/3+kπ,π/6+kπ],k∈Z 上单调递减
当k=0时函数的减区间为[-π/3,π/6]
当k=1时函数的减区间为[2π/3,7π/6]
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y=cos(-2x+π/3)=cos(2x-π/3)
函数在(2x-π/3)∈[-π+2kπ,2kπ],k∈Z (将2x-π/3看作一个整体)
即x∈[-π/3+kπ,π/6+kπ],k∈Z 上单调递减
当k=0时函数的减区间为[-π/3,π/6]
当k=1时函数的减区间为[2π/3,7π/6]
所以y=cos(-2x+π/3)在[0,π])单调递减区间为
[0,π/6]和[2π/3,π]
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