求函数y=√2sin(2x-π)cos[2(x+π)]的单调区间
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求函数y=√2sin(2x-π)cos[2(x+π)]的单调区间
∵y=-√2sin(π-2x)cos(2π+2x)=-√2sin2xcos2x=-(√2/2)sin4x.
∴当2kπ-π/2<4x≦2kπ+π/2时,y单调递增,
当2kπ+π-π/2<4x≦2kπ+π+π/2时,y单调递减.
由2kπ-π/2<4x≦2kπ+π/2,得:kπ/2-π/8<x≦kπ/2+π/8.
由2kπ+π-π/2<4x≦2kπ+π+π/2,得:kπ/2+π/8<x≦kπ/2+3π/8.
∴原函数的单调递增区间是(kπ/2-π/8,kπ/2+π/8],
单调递减区间是(kπ/2+π/8,kπ/2+3π/8].其中k为整数.
∴当2kπ-π/2<4x≦2kπ+π/2时,y单调递增,
当2kπ+π-π/2<4x≦2kπ+π+π/2时,y单调递减.
由2kπ-π/2<4x≦2kπ+π/2,得:kπ/2-π/8<x≦kπ/2+π/8.
由2kπ+π-π/2<4x≦2kπ+π+π/2,得:kπ/2+π/8<x≦kπ/2+3π/8.
∴原函数的单调递增区间是(kπ/2-π/8,kπ/2+π/8],
单调递减区间是(kπ/2+π/8,kπ/2+3π/8].其中k为整数.
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