设f(x)=6cos^2x-根号3sin2x,求f(x)的单调增区间和单调减区间
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/19 16:42:47
x){n_FYr~qQ;o7.3y $|Vީ/64?]4iϮӷ<i$S0;*
I`Y[cmcPQ,Tq#PNèB|Xf0;Oh}F 10nu=(tjRU6FRVصp zgȩ@c7<\W
b 7 ́*f Pbzff$ zںe>3ć棎(VX:<pW)`8]PagSغI f.
设f(x)=6cos^2x-根号3sin2x,求f(x)的单调增区间和单调减区间
设f(x)=6cos^2x-根号3sin2x,求f(x)的单调增区间和单调减区间
设f(x)=6cos^2x-根号3sin2x,求f(x)的单调增区间和单调减区间
f(x)=6cos^2x-根号3sin2x
=3+3cos2x-√3sin2x
=3+2√3cos(2x+π/6),
它的减区间由2kπ
f(x)=6cos²x-√3sin2x
=3(cos2x+1)-√3sin2x
=3cos2x-√3sin2x+3
=2√3cos(2x+π/6)+3
增区间:2kπ-π≤2x+π/6≤2kπ
kπ-7π/12≤x≤kπ-π/12
即增区间是:[kπ-7π/12,kπ-π/12],k...
全部展开
f(x)=6cos²x-√3sin2x
=3(cos2x+1)-√3sin2x
=3cos2x-√3sin2x+3
=2√3cos(2x+π/6)+3
增区间:2kπ-π≤2x+π/6≤2kπ
kπ-7π/12≤x≤kπ-π/12
即增区间是:[kπ-7π/12,kπ-π/12],k∈Z
减区间:2kπ≤2x+π/6≤2kπ+π
kπ-π/12≤x≤kπ+5π/12
即减区间是:[kπ-π/12,kπ+5π/12],k∈Z
收起
设f(sin x)=3-cos 2x,则f(cos x)=?
设函数f(x)=cosωx(根号3×sinωx+cosωx)其中0
已知F(X)=根号3COS^2 X+SIN XCOS X-2SIN X*SIN(X-π/6),求F(X)的最大值
设函数f(x)=sin(x+π/3)+2sin(x-π/3)-根号3cos(2π/3-x)求f(π/6) f(π/3)的值.. 求过程,求速度..
设函数f(x)=根号3 sin x cos x+cos平方x+a.写出函数f(x)的最小正周期极单调递减区间;
设函数f(x)=a*b,其中向量a=(2cos x,1),b=(cos x,根号3 sin 2x),x属于(-30,30度),f(s)=3/4,求cos(2x)
设函数f(x)=(sinθ/3)x^3+((根号3)cosθ/2)x^2+tanθ,则f'(π/4)=
函数f(x)=根号3sinωx+cosωx(ω>0)怎样变为f(x)=2sin(ωx+π/6)
设f(x)=根号1-x化简f(sin2)+f(sin(-2))1.设f(x)=根号1-x化简f(sin2)+f(sin(-2))2.已知tanθ=2,求2sinθ的平方+3sin2θ+13.若sinθ:sinθ/2=8:5求cosθ
已知函数f(x)=根号3sin(2x+fai)-cos(2x+fai)(0
函数f(x)=sin(2x+q)+根号3cos(2x+q)(0
已知函数f(x)=sin(2x+α)+根号3cos(2x+α)(0
已知f(x)=cos^4x-2根号3sinxcosx-sin^4x 化简
化解三角函数F(x)=sin平方x+根号3sinxcosx+2cos平方x
设函数f(x)=sin²x+2sin2x+3cos²x 化简
设f(x)=(sin^2(6π+x)+cosx-2cos^3(3π+x)-3)/2+cos^2(x-4π)-cos(-x)设f(x)=(sin^2(6π+x)+cosx-2cos^3(3π+x)-3)/2+cos^2(x-4π)-cos(-x)求f(π/3)的值
f(x)=2sin^2(π/4+x)+根号3(sin^2x+cos^2x)f(x)=2sin^2(π/4+x)+根号3(sin^2x-cos^2x) 符号打错了
设fx=1/2cos^2x+根号3sinxcosx+3/2sin^2x