设x∈(0,π/2),如何求sin^2xcos^2x+2/sin^2xcos^2x-2的最小值.
设x∈(0,π/2),如何求sin^2xcos^2x+2/sin^2xcos^2x-2的最小值.
求不定积分(1/sin^2xcos^2x)dx
求不定积分,∫sin^2xcos^2x dx
sin^2x+cos^2x)(sin^4x-sin^2xcos^2x+cos^4x) =sin^4x-sin^2xcos
求函数f(x)=(sin^4x+cos^4x+sin^2xcos^2x)/(2-sin2x)的最小正周期、最大值和最小值
求函数fx=sin^4x+cos^4x+sin^2xcos^2x/2-2sinxcosx的最小正周期,最大值和最小值
求函数y=2sin xcos x+2sin x+2cos x+4的值域
求sin^4x+cos^4x+4sin^2xcos^2x-1的最小正周期及值域.
(1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x
化简cos^4x+sin^2xcos^2x+sin^2x
求函数f(x)=(sin^4x+cos^4x+sin^2xcos^2x)/2
求函数f(x)=(sin⁴x+cos⁴x+sin²xcos²x)/(2-si