y=sin(2x-π/6) x∈（0,π ）的值域y=cos(x+π/3) x∈（-π/2，π/2 ）的值域

判断单调性 y=sin(x-π/6) y=sin(-2x+π/3) y=2sin(2x+π/6）x∈[0,π/2],求值域 函数y=2sin(x+π/6),x∈【0,π/2】,值域是 求y=sin(2x+π/6)x∈(0,π/3)的值域 函数y=sin(x+π/6)（x∈[0,π/2])的值域是 函数y=2sin(x-π/6)(x∈[0,π])的值域请求详解 y=sin(x+π/6)+sin(x-π/6)+cosx ,x∈[-π/2,π/2]的值域 函数y=2sin(2x-π/6)(0 已知sinx=3/5,x∈(π/2,π),求【sin(x+y)+sin(x-y)】/【cos（x+y）+cos（x-y）】的值 y=sin(x-π/6) x∈【0,π】函数y=sin（x-π/6）,x∈【0,π】的值域是 5cos（2x-y）+7cosy=0 ,tan（x-y）tanx=?y=sin²（x+π/4)-sin²（x-π/4),x∈（π/6,π/3）值域 y=12sin(2x+π/6)+5sin(π/3-2x)最大值 y=12sin(2x+π/6)+5sin(π/3-2x)的最大值 函数y=2sin(π/3-X)+sin(π/6+X)的最大值 解微分方程：x(y´+1)+sin(x+y)=0, y(π/2)=0 y=sin(2x+π/6),x∈[π/4,3π/4]求值域 求函数的值域y=sin(2x+π/3),x∈(-π/6,π) 函数y=sin(x-6分之π),x∈【0,π】的值域为