2cos平方x-sin2x单调递增区间

y=sin平方x+sin2x+3cos平方x=1+sin2x+2cos^2x=1+sin2x+2cos^2x-1+1=sin2x+cos2x+2=√2sin(2x+π/4)+2当sin(2x+π/4)

f(x)=2cos平方x+sin2x+a=2cos平方x-1+1+sin2x+a=cos2x+1+sin2x+a=sin2x+cos2x+a+1=√2(sin2xcosπ/4+cos2xsinπ/4)

y=2cos平方x+sin2x/1+tanx=2(sinx+cosx)cosx/[(cosx+sinx)/cosx]=2cos^2x=cos2x+10

（1+sinx）/cosx*sin2x/(2cos²（π/2-x/2）=（1+sinx）/cosx*sin2x/(2sin²（x/2）)=（1+sinx）/cosx*2sinxco

原式=(1+sinx)/cosx*2sinxcosx/[1+cos(π/2-x)]=(1+sinx)/cosx*2sinxcosx/(1+sinx)=2sinx

f(x)=2cos²x+√3sin2x=(2cos²x-1)+1+√3sin2x=cos2x+√3sin2x+1=2sin(2x+π/6)+1函数f(x)最小正周期T=2π/2=π

化简得：y=sin（2x+π/3）+根号3/2周期是π,增区间是（kπ-5π/12,kπ+π/12)(k属于z)

化简,得到：f(x)=2sin(2x＋π/6)－1利用：函数y=sinx的减区间是：[2kπ＋π/2,2kπ＋3π/2],则函数f(x)的减区间是：2kπ＋π/2≤2x＋π/6≤2kπ＋3π/2解得：

y=√3cos²x+1/2sin2x=√3/2(cos2x+1)+1/2sin2x=√3/2cos2x+1/2sin2x+√3/2=sin(π/3)cos2x+cos(π/3)sin2x+√

f(x)=sin2x-2√3(cosx)^2+√3=sin2x-√3(1+cos2x)+√3=sin2x-√3cos2x=2sin(2x-π/3)π/4=再问：π/6=

f(x)=cos2x+√3sin2x+a+1=2sin(2x+π/6)+a+1单调增区间2kπ-π/2

∵cos2x=2cos²x-1,∴cos²x=1/2(cos2x+1)∴原式=1/(1/2cos2x+sin2x+1/2)=1/[根号5/2sin(2x+ψ)+1/2](tanψ=

sinx/cosx=tanx=2sinx=2cosxsin²x=4cos²x因为sin²x+cos²x=1所以cos²x=1/5sin2x=2sinx

【1】f(x)=(1-sin2x)/(2cosx^2)f(x)=(1-sin2x)/(1+cos2x)1+cos2x≠0cos2x≠-12x≠π+2kπx≠π/2+kπ函数定义域:{x|x≠π/2+k

cos^2x=cos2x+1所以原式等于二分之根号三倍的（cos2x+1)+1/2sin2x化简得sin(2x+60)+二分之根号三最大值就是1+二分之根号三

f(x)=根号3sin2x+2cos平方x+1=√3sin2x+cos2x+2=2sin(2x+π/6)+2所以当2kπ-π/2

 再问：那f(x)的最值，还有单调区间再问：额，好像会了。谢谢

f(x)=2cos^2x+√3sin2x=cos2x+1+√3sin2x=2(sin2xcosπ/6+cos2xsinπ/6)+1=2sin(2x+π/6)+1f(C)=22sin(2C+π/6)+1

f(x)=根号3cos^2x+1/2sin2x=根号3/2*cos2x+1/2sin2x+根号3/2=sin(2x+π/3)+根号3/2y=sinx递减区间[2kπ+π/2,2kπ+3π/2]y=si