方程|tan2x|=sin(x-π/4)在（-π,π）实数解的个数有几个?

解方程4sinx cox=3tan2x - 6tan2x sin^2x,0≦x≦2pi 方程|tan2x|=sin(x-π/4)在（-π,π）实数解的个数有几个? 求证(tanxtan2x/tan2x-tanx)/(tan2x-tanx)+√3（sin^2x-cos^2x）=2sin(2x-π/3) 求证：（1-2sin2xcos2x）/(cos^2x-sin^2x)=(1-tan2x)/(1+tan2x) (1-2sin2xcos2x)/(cos^2 2x-sin^2 2x)=(1-tan2x)/(1+tan2x) 求证 (1-2sinxcosx)/(cos^2x-sin^2x)=(1-tan2x)/(1+tan2x) 函数y=(1-tan2x)/(1+tan2x)的最小正周期是?还有一道：函数y=sin^4 x+cos^2 x的最小正周期是? 求函数y=√tan2x-1的定义域tan2x－1≥0tan2x≥1则：kπ＋π/4≤2x 已知tan2x=-2 π/2<2x<π求[2cos^2(x/2)-sinx-1]/根号2sin(π/4+x) 求证：(sin x+cos x)(1-tan x)=(2sin x)/(tan2x) (cos^2 2x－sin^2 2x)/(1－tan2x)=(1+2sin2xcos2x)/(1+tan2x) 求证:{[1-2sin2xcos2x]/[cos^2(2x)-sin^2(2x)]}={(1-tan2x)/(1+tan2x)} 已知x属于(0,2/π),tanx=2/1,tan2x和sin(2x+3/π)的值 三角函数数学题已知2sin^2+sinXcosX-4sinX-2cosX=0,求（1+sin2X)/(1-cos2X)(1-tan2X)应该是：已知2sin^2X+sinXcosX-4sinX-2cosX=0,求（1+sin2X)/(1-cos2X)(1-tan2X) 证1-tan2x/1+tan^2x=cos^2x-sin^2x 函数y=(cos^2x-sin^2x)tan2x的最小正周期为____________ 已知sin^2x+sin2x*sinx+cos2x=1,x属于(0,派/2),求tan2x 函数y=(cos^2x-sin^2x)*tan2x的最小正周期是 .