若f(x)=4cos^2x+4根号3sinxcosx-2,x属于R,求周期,求单调增区间,当x[-派/3,派/4]时f(

4cos^2X=1-2sinXcosX
1-4cos^2X--2sinXcosX=0
1-cos^2X-3cos^2X-2sinXcosX=0
sin^2X-2sinXcosX-3cos^2X=0
(sinX-3cosX)(sinX+cosX)=0
所以
sinX-3cosX=0 或 sinX+cosX=0
tanX=3 或 -1
所以
X=arctan3 或 -π/4±nπ（n为整数）

y=2(1+cos2x) + 2√3sin2x - 2 = 2(cos2x + √3sin2x)=4sin(2x+π/6) （1）周期T=2π/2=π （2）当2x+π/6=π/2+2kπ,即x=π/6+kπ,k∈Z时,y有最大值4 （3）单调增区间：-π/2+2kπ≤2x+π/6≤π/2+2kπ,即-π/3+kπ≤x≤π/6+kπ,k∈Z 即单调增区间是：[-π/3+kπ,π/6+kπ] (4)对称轴2x+π/6=π/2+kπ,所以x=π/3+kπ/2

f(x)=4cos^2x-4√3sinxcosx-1
=8cosx（1/2cosx-√3/2sinx）-1
=8cosx（sinπ/6cosx-cosπ/6sinx）-1
=8cosxsin（π/6-x）-1
=4（sinπ/6-sin（2x-π/6）-1
=2-4sin（2x-π/6）-1
=1-4sin（2x-π/6）
（1）最小正周期T=2π/ω=π
（2）单调递增区间
2Kπ+π/2≤2x-π/6≤2Kπ+3π/2
2Kπ+2π/3≤2x≤2Kπ+5π/3
Kπ+π/3≤x≤Kπ+5π/6
（3）2x-π/6=Kπ+π/2
2x=Kπ+2π/3
x=Kπ/2+π/3
对称轴为x=Kπ/2+π/3

(1) 4cos²x-4sinx=1
4(1-sin²x)-4sinx=1
4sin²x+4sinx-3=0
(2sinx-1)(2sinx+3)=0
sinx=1/2 (sinx=-3/2

y=4cos^2x-2+4√3sinxcosx
=2(2cos^2x-1) +4√3sinxcosx
=2cos2x+2√3sin2x
=4[(√3/2)sin2x+1/2cos2x]
=4sin(2x+π/6)
函数的最小正周期是2π/2=π.
函数的最大值是4,此时2x+π/6=2kπ+π/2,k∈Z.
则x= kπ+π/6,k∈Z.

sin2x=(2tanx)/(1+tan^2x)=4/5
(1)sinxcosx=1/2sin2x=2/5
(2)(2sin^2x-3sinxcosx-4cos^2x )/(sinxcosx)=2tanx-3-4cotx=-1
2sin^2x-3sinxcosx-4cos^2x =-sinxcosx=-2/5

sin^2x+cos^2x=1
sin^2x/cos^2x=4
sin^2x=0.8
cos^2x=0.2
代入就行了!

f(x)=4cos^4x+4cos^2x-3/cos2x=4cos^4x+4cos^2x-3/(2cos^2x-1)
因为2cos^2x-1不为0,故定义域为{xx不等于土pi/4+2k*pi,k为整数)
令cos^2x=t,t属于[0,1)
那么原式为f(t)=4t^2+4t-3/(2t-1)
求导,f'(t)=8t+4+3/(2t-1)^2
因为t属于[0,1),故f'(t)恒大于0,所以f(t)单调递增,其值域为[3,5)