求函数y=sin(-2x+6/π)的单调递减区间
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求函数y=sin(-2x+6/π)的单调递减区间
求函数y=sin(-2x+6/π)的单调递减区间
求函数y=sin(-2x+6/π)的单调递减区间
y=sin(-2x+π/6)=sin[π-(-2x+π/6)]=sin(2x+5π/6)
则递减区间是:2kπ+π/2≤2x+5π/6≤2kπ+3π/2
得:kπ-π/6≤x≤kπ+π/3
即减区间是:[kπ-π/6,kπ+π/3],k∈Z
y=sin(-x)周期为2π在-π~π上的单调递减区间为-π/2
2kπ-π/2<(2x-6/π)<2kπ+π/2
即2kπ-π/2+6/π<2x<2kπ+π/2+6/π
kπ-π/4+3/π
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y=sin(-x)周期为2π在-π~π上的单调递减区间为-π/2
2kπ-π/2<(2x-6/π)<2kπ+π/2
即2kπ-π/2+6/π<2x<2kπ+π/2+6/π
kπ-π/4+3/π
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y=sin(-2x+6/π)
=-sin(2x-π/6)
也就是求
sin(2x-π/6)的单调增区间
2x-π/6∈[2kπ-π/2,2kπ+π/2]
x∈[kπ-π/6,kπ+π/3]
所以
函数y的单调减区间为
[kπ-π/6,kπ+π/3]y=sin(-2x+6/π)又不是奇函数,怎么可以提出负号呢?不是奇函数就不能提了吗 ...
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y=sin(-2x+6/π)
=-sin(2x-π/6)
也就是求
sin(2x-π/6)的单调增区间
2x-π/6∈[2kπ-π/2,2kπ+π/2]
x∈[kπ-π/6,kπ+π/3]
所以
函数y的单调减区间为
[kπ-π/6,kπ+π/3]
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