2√3sinxcosx+2cos²x-1 =√3sin2x+cos2x =2sin(2x+π/6)

(1)f(x)=sin2x+cos2x+1
=√2(sin2x *√/2+cos2x *√2)+1
=√2(sin2x cosπ/4+cos2x cosπ/2)+1
=√2sin(2x+π/4)+1
(2) 当x∈[0,π/2]
2x+π/4∈[π/4,3π/4]
sin(2x+π/4)∈[√2/2,1]
所以 y∈【2,√2+1】

f（x）=2√3sinxcosx+2cos²x-1（x∈R）
f(x) = 2√3sinxcosx+2cos²x-1
= √3 * ( 2sinxcosx) + (2cos²x-1)
= 根号3 sin2x + cos2x
= 2 (sin2xcosπ/6+cos2xsinπ/6)
=2 sin(2x+π/6)
最小正周期：2π/2 = π
在区间[0,π/2]
x ∈[0,π/2]
2x ∈[0,π]
2x+π/6 ∈[π/6,π+π/6]
2x+π/6∈[π/6,π/2]时单调增；
2x+π/6∈[π/2,π+π/6]时单调减
2x+π/6=π/2时有最大值,2sinπ/2=2
2x+π/6=π/6时,f(x)=2sinπ/6=1
2x+π/6=π+π/6时,f(x)=2sin(π+π/6)=-1
所以最小值-1,最大值2
f(x0）=6/5
2 sin(2x0+π/6)=6/5
sin(2x0+π/6)=3/5.(1)
x0∈[π/4,π/2]
2x0∈[π/2,π]
2x0+π/6 ∈[2π/3,7π/6]
cos(2x0+π/6) = - 根号{-sin^2(0+π/6)} = - {1-(3/5)^2} = -4/5.(2)
由（1）：sin2x0cosπ/6+cos2x0sinπ/6=3/5
根号3/2 sin2x0 + 1/2 cos2x0 = 3/5
根号3 sin2x0 + cos2x0 = 6/5 .(3)
由（2）：cos2x0cosπ/6 - sin2x0sinπ/6 = -4/5
根号3/2 cos2x0 - 1/2 sin2x0 = -4/5
根号3 cos2x0 - sin2x0 = -8/5 .(4)
(3)+(4)*根号3得：
cos2x0+3cos2x0 = 6/5 -8根号3/5
4cos2x0= 6/5 -8根号3/5
cos2x0= 3/10 - 2根号3 /5
求cos2x0

f(x)=2sinxcosx+2cos²x
=sin2x+cos2x+1
=√2sin(2x+π/4)+1
最大值√2+1
T=π

f(x)=2sinxcosx+2cosx的平方-1=sin2x+cos2x=√2sin(2x+π/4)
f(x)的周期=π
f(x)的最大值=√2
单调递增区间[-3π/8+kπ,π/8+kπ]

已知函数f(x)=2√3sinx×cosx+2cos²x-1(x∈R)
求函数f(x)的最小正周期及在区间[0,π/2]上的最值
若f(x0)=6/5,x0∈[π/4,π/2],求cos2x0的值
(1)解析：∵函数f(x)=2√3sinxcosx+2cos²x-1=√3sin2x+cos2x=2sin(2x+π/6)
∴f(x)的最小正周期为π
f(0)=2sin(π/6)=1,f(π/2)=2sin(π+π/6)=-1,f(π/6)=2sin(π/3+π/6)=2
在区间[0,π/2]上的最大值为2,最小值为-1
(2)解析：∵f(x0)=2sin(2x0+π/6)=6/5==>sin(2x0+π/6)=3/5
∴sin2x0*√3/2+cos2x0*1/2=3/5
√3sin2x0+cos2x0=6/5
sin2x0=2√3/5-√3/3cos2x0==> (sin2x0)^2=(2√3/5-√3/3cos2x0)^2=12/25-4/5(cos2x0)+1/3(cos2x0)^2
代入(sin2x0)^2+(cos2x0)^2=1
4/3(cos2x0)^2-4/5(cos2x0)^2-13/25=0
解得cos2x0=3/10-2√3/5或cos2x0=3/10+2√3/5

等我给你写,

由2sinxcosx=sin2x 2cos²x-1 =cos2x 这是公式 化简得√3sin2x+cos2x 再有化为2（（√3/2）sin2x+（1/2）cos2x）=2（cosπ/6sin2x+sinπ/6cos2x）=22sin(2x+π/6)

2√3sinxcosx+2cos²x-1
=√3*2sinxcosx+2cos²x-1
正弦二倍角公式得
2sinxcosx=sin2x
余弦二倍角公式得
2cos²x-1=cos2x
∴原式
=√3sin2x+cos2x
接下来
=2(√3/2*sin2x+1/2*cos2x)
=2(sin2xcos30°+cosxsin30°)
=2sin(2x+30°)

f(x)=2√3sinxcosx+2cos²x-1
=√3sin2x+cos2x
=2(√3/2sin2x+1/2cos2x)
=2(sin2xcosπ/6+cos2xsinπ/6）
=2sin(2x+π/6)
(1)
对称轴：2x+π/6=2kπ+π/2
2x=2kπ+π/3
x=kπ+π/6；k∈Z
闭区间【0,π/2】
当x=π/2时；函数有最小值=-2sin(π/6)=-1
当x=π/6时,函数有最大值=2sin(π/2)=-2
(2)2sin(2x0+π/6)=6/5
sin(2x0+π/6)=3/5
sin2x0cosπ/6+cos2x0sinπ/6=3/5
√3sin2x0+cos2x0=6/5 （1）
x0∈[π/4,π/2]
2x0∈[π/2,π]
cos2x0≤0
2x0+π/6∈[2π/3,7π/6]
因为：sin(2x0+π/6)=3/5>0
所以：2x0+π/6∈[2π/3,π)
所以：cos(2x0+π/6)

(1)
∵f(x)=2sinxcosx+2cos²x=2sinxcosx+(2cos²x-1)+1=sin(2x)+cos(2x)+1=(√2)sin(2x+π/4)+1
∴T=2π/2=π.
令2x+π/4=3π/2+2kπ,则x=5π/8+kπ(k∈Z).
即当x=5π/8+kπ(k∈Z)时,f(x)有最小值1-√2.
(2)
∵f(x)=(√2)sin(2x+π/4)+1
∴f(x+π/8)=(√2)sin[2(x+π/8)+π/4]+1=(√2)sin(2x+π/2)+1=(√2)cos(2x)+1
∴g(x)=f(x+π/8)-1=[(√2)cos(2x)+1]-1=(√2)cos(2x)
∵x∈[-π/6,π/3]
∴2x∈[-π/3,2π/3]
由余弦函数图像知：最大值是1
则g(x)=(√2)cos(2x)的最大值就是√2.
∵g(x)√2
∴a>2+√2.
∴a的取值范围是(2+√2,+∞).

以f(x)=2sin x来看,离原点最近的对称中心为(0,0),
现在f(x)=2sin(2x十兀/6)只是相当于x变为了2x十兀/6,即2X十兀/6=0,故x=-兀/12,即坐标为(-兀/12,0)

已知函数f（x)=2sinxcosx+2cosx,求f(4分之π）的值
∵f（x)=2sinxcosx+2cosx
=sin2x+2cosx,
∴f(4分之π）
=sinπ+2cos(π/4)
=1+2*√2/2
=1+√2.

f(x)=(2cos^2(x)-1)+2√3sinxcosx
=cos2x+√3sin2x
=2sin(2x+π/4)
T=π

f(x)=2sinxcosx+2cosx的平方-1 =sin(2x)+cos(2x) =根号2[根号2/2·sin(2x)+根号2/2·cos(2x)] =根号2[cos(π/4)·sin(2x)+根号2/2·cos(2x)] =根号2sin(2x+π/4) =根号2sin(2π+2x+π/4) =根号2sin[2(π+x)+π/4]； 所以f(x)的最小正周期是π.

f(x)=2cosx(sinx-cosx)+1
=2sinxcosx-2(cosx)^2+1
=2sinxcosx-[2(cosx)^2-1]
=sin2x-cos2x
=√2(√2/2*sin2x-√2/2*cos2x)
=√2(sin2xcosπ/4-cos2xsinπ/4)
=√2sin(2x-π/4)
（1）
f(x)=√2sin(2x-π/4)
∴函数f(x)的最小正周期:
T=2π/2=π
（2）2sin(2x0+30°)=6/5
则sin(2x0+30°)=3/5
cos(2x0+30°)=-4/5
所以cos2x0=（3-4√3）/10
不懂的欢迎追问,

已知函数f(x)=sin^2 x √3sinxcosx 2cos^2 x,求f(x)最小正周期和有以上三式解得a=-4,b=2,c=8 3.设等差数列｛An｝的首项A1及公差

(1)
∵f(x)=2sinxcosx+2cos²x=2sinxcosx+(2cos²x-1)+1=sin(2x)+cos(2x)+1=(√2)sin(2x+π/4)+1
∴T=2π/2=π.
令2x+π/4=3π/2+2kπ,则x=5π/8+kπ(k∈Z).
即当x=5π/8+kπ(k∈Z)时,f(x)有最小值1-√2.
(2)
∵f(x)=(√2)sin(2x+π/4)+1
∴f(x+π/8)=(√2)sin[2(x+π/8)+π/4]+1=(√2)sin(2x+π/2)+1=(√2)cos(2x)+1
∴g(x)=f(x+π/8)-1=[(√2)cos(2x)+1]-1=(√2)cos(2x)
∵x∈[-π/6,π/3]
∴2x∈[-π/3,2π/3]
由余弦函数图像知：最大值是1
则g(x)=(√2)cos(2x)的最大值就是√2.
∵g(x)√2
∴a>2+√2.
∴a的取值范围是(2+√2,+∞).

1f(x)=2sinx乘cosx+2cos²x-1=sin2x+cos2x=√2(√2/2sin2x+√2/2cos2x)=√2sin(2x+π/4)f(x)最大值为√22∵点p(-3,4)在角α的终点上x=-3,y=4,r=5∴cosa=x/r=-3/5f(a+π/8)=√2sin[2(a+π/8)+π/4]=√2sin(2a+π/...