y=2sin^2x-sinxcosx+3cos^2x求最小正周期,单调增区间,在[0,π/2]上的值域

函数y=sinxcosx-3sin^2(x)的最大值 y=(cos^2x)/(sinxcosx-sin^2x),(0 求y=sin^2x+2sinxcosx的周期 y=sin^2x+2sinxcosx的值域 y=sin^2x-sinxcosx+1/2的值域 求y=sin^2(x)-sinxcosx+2的值域 求y=sin平方x+2sinxcosx的周期 求y=sinxcosx+sin^2x,0 y=1/2cos^2x+sinxcosx+3/2sin^2x 求单调区间 y=2sin^2x-sinxcosx+3cos^2x 求值域 y=sin'2x+根号3sinxcosx+2cos'2x求值域和周期 y=sin^2x+2sinxcosx+3cos^2x的值域? y=sin^2x-根号3sinxcosx-cos^2x化简 y=1/2cos²x+sinxcosx+3/2sin²x 求化简详程. y=sin^2x+sinxcosx+3cos^2x的最值 y=sin^2X+2sinXcosX+3COS^X谢谢了, y=sin^2x+3sinxcosx+5cos^2x值域 函数y=sin²x+2sinxcosx-3cos²x求周期?求最大值?