f(x)=根号3sin2x=2cos^2x m在区间[0,π 2]
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/08 15:41:48
注:题中f(x)=(√3)sin2x-2sin²x-1 解.(1)f(x)=(√3)sin2x-2sin²x-1 =(√3)sin2x+cos2x-2 =2sin(2x+π/
f(x)=1+cos2x+根号3sin2x+a=2sin(2x+π/6)+a+11、若f(x)max=2,则sin(2x+π/6)=1,即2+a+1=2,得a=-12、正弦的单调减区间在第二和第三象限
f(x)=2根号3sin方x+sin2x+根号3=根号3(2sin方x+1)+sin2x=根号3(1-cos2x+1)+sin2x=2根号3-根号3cos2x+sin2x=2sin(2x-60度)+2
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=
已知函数f(x)=根号3sin2x+cos2x+21求f(x)的最大值及f(x)取得最大值时自变量x集合f(x)=根号3sin2x+cos2x+2=2[(根号3/2)sin2x+(1/2)cos2x]
ab=√3(sin2x)^2+cos2xsin2x=(√3/2)(1-cos4x)+(1/2)sin4x=(√3/2)+sin4xcos兀/3-cos4xsin兀/3=(√3/2)+sin(4x-兀/
一样是先化简的问题f(x)=2cos^2x+√3*sin2x-1=2(cosx)^2-1+√3*sin2x=cos2x+√3*sin2x=2(1/2*cos2x+√3/2*sin2x)=2(sinπ/
f(x)=2(cosx)^2+√3(sin2x)=1+2sin[2x+(π/6)].(1)∵x∈[0,π],∴2x+(π/6)∈[π/6,13π/6],∴递增区间[0,π/6]和[2π/3,π].(2
1,f(x)=√3sin2x+cos2x+1-m=2sin(2x+π/6)+1-m.若0再问:2A+π/6=π/2,为什么不行再答:sin(2A+π/6)=1/2,怎么可能是2A+π/6=π/2呢?再
f(x)=sin2x+√3cos2x=2(1/2sin2x+√3/2cos2x)=2sin(2x+π/3)f(x)的最大值=2,最小值=-2f(x)的单调区间:∵2kπ-π/2≤2x+π/3≤2kπ+
1).f(x)=(√3)sin2x+cos2x+2=2sin(2x+π/6)+2单增区间[-π/3+kπ,π/6+kπ](2)f(C)=2sin(2C+π/6)+2=3,故C=π/3a/b=sinA/
(1)f=[根号3]/2sin2x-cos^2(x)-1/2=√3/2*sin2x-1/2(1+cos2x)-1/2=√3/2sin2x-1/2cos2x-1=sin(2x-π/6)-1∵x∈[-π/
1∵点p(1,-根号3)在角a的终边上,∴tana=-√3∴f(a)=√3sin2a-2sin²a=(2√3sinacosa-2sin²a)/(sin²a+cos
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
(1)、f(x)的定义域为sinx≠0,即x≠kπ;如果定义x=kπ时f(x)等于-1,可将定义域扩大至整个实数域;f(x)=[2√3sinxcosx-2cos²x]+1=√3sin2x-c
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
f(x)=(根号3/2)sin2x-cos^2x+1/4=(根号3/2)sin2x-(2cos^2x-1)/2+3/4=(根号3/2)sin2x-(cos2x)/2+3/4=sin((2x-π/6))