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函数y=sin(2x%3/π)的单调增区间是

解: y=3sin(π/3-2x) =-3sin(2x-π/3) 在2x-π/3∈(2kπ+π/2,2kπ+3π/2)时单调增 故单调增区间是: x∈(kπ+5π/12,kπ+11π/12) k∈Z 如仍有疑惑,欢迎追问。 祝:学习进步!

T=2π/2=π,增区间:-π/2+2kπ

对于函数y=sinx来说,它的减区间就是[π/2+2kπ,3π/2+2kπ],因而: 令π/2+2kπ

原题是:函数y=sin(π/3-x/2),x∈[0,4π],求函数的单调递减区间. y=sin(π/3-x/2)=-sin(x/2-π/3) =sin(x/2-π/3+π)=sin(x/2+2π/3) 即y=sin(x/2+2π/3) x∈[0,4π],x/2+2π/3∈[2π/3,2π+2π/3] 当2π/3≤x/2+2π/3≤2π/2或2π+π/2≤x/2+2π/3≤2π+2π/3 即0≤x/≤5π/3或...

是2x-π/6吧 y=-3sin(2x-π/6)=3sin(2x+5π/6) 和书上求区间的不同方法 根据正弦函数的图像,对称轴π/2+2kπ的前半个周期是增区间 所以y=3sin(2x+5π/6)的对称轴是 2x+5π/6=π/2 x=-π/6+kπ 所以增区间是[-2π/3+kπ,-π/6+kπ] 可追问 望采纳 附加图

令2kπ-π/2

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(2x+π/4)的周期为π,那么y=|sin(2x+π/4)|的周期为π/2,所以只要计算一个零点即可,令2x+π/4=0得x=-π/8,在[-π/8+kπ/2,π/8+kπ/2]为单调增区间

∵函数f(x)在区间π/6到2π/3上函数值从1减小到-1 ∴T/2=2π/3-π/6=π/2,∴T=π 由T=2π/w=π==>w=2 ∵x=π/6时,f(x)取得最大值1 ∴sin(2*π/6+φ)=1 ∴2*π/6+φ=kπ+π/2,k∈Z ∵|φ|

解: (1)y=sin(π/3+4x) 周期为T=2π/w=2π/4=π/2 单调区间: 2kπ-π/2≤π/3+4x≤2kπ+π/2 (k∈Z)得 kπ/2-5π/24≤x≤kπ/2+π/24 即单调递增区间为[kπ/2-5π/24,kπ/2+π/24] (k∈Z) 2kπ+π/2≤π/3+4x≤2kπ+3π/2 (k∈Z)得 kπ/2+π/24≤x≤kπ/2+7π/24 即单调递减区...

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