sin(2x+π/6)若x属于【-π/6,π/3】,求f(x)的值域为什么我最后算出来-π/6

若x属于(0,π/2),且6 cosx=5 tan X,则sin X 求下列函数周期 y=3sin(x/4) x属于Ry=2sin(2x-π/6) y=2sin(x/2+π/6),x属于R y=-6sin(2x-π/3）,x属于R 写出振幅,周期,初相 f(x)=2sin(2x-π/6) 若x属于【0,π】 求y的单调增区间 y=sin(2x+π/6),x属于[π/6,3π/4],求值域 y=2sin[(π/6)-2x]的单调递增区间 x属于[0,π] 函数y=sin（x+π/6）,x属于【0,π/2】的值域是? 1.y=sin(1/2x+π/6),x属于[0,π/3] 2.y=-cos(3x-π/3),x属于[-π/3,π/3] f(x)=2sin(2x+π/6),若x属于[0,π/2] ,求函数f(x)的值域 急!数学小题!已知f(x)=根号3*sin2x-2sin^2x.若x属于[-π/6,π/3],求f(x)的值域 sin(x+π/6)=-√2/2,x属于-π/6,11π/6,求x f(x)=2sin(2x+π/6) x属于[π/12,π/2] 求值域 sin^2(π/3-x)+sin^2(π/6+x) 函数y=2sin((π/3)-x)-cos((π/6)+x) (x属于R）的最小值是 当x属于[0,π/2]时,求函数f(x)=sin(π/6-x)-cosx的值域 已知函数f(x)=sin(2x+π/6)+sin(2x—π/6)—cos2x+a(a为实数,属于R)化简得f（x）=2sin（2x－π/6）＋a若x属于【π/4,π/2】时,F（x)的最小值是-2,求a 函数f(x)=cos^4x-2sinxcosx-sin^4x,x属于[0,π/2],求f(x)的最大值,最小值函数f(x)=cos^4x-2sinxcosx-sin^4x,x属于[0,π/2],求f(x)的最大值,最小值,求f(x)的单调区间。若π/4小于等于v=x小于等于π/3，求函数y={sin(x+π/6) 已知sin（π/4+x）sin（π/4-x）=1/6,x属于（π/2,π）,求sin4x的值