sinx^2-2sinxcosx+2cosx^2=1/2,则Tanx=?请给具体过程.

答：
f(x)=2√3sinxcosx+2cos²x-t
=√3sin2x+cos2x+1-t
=2 [ (√3/2)sin2x+(1/2)cos2x ] +1-t
=2sin(2x+π/6)+1-t
0

根号3*2sinxcosx+2cos²x-1+1
=根号3*sin2x+cos2x+1
=2(根号3/2sin2x+1/2cos2x)+1
=2sin(π/3+2x)+1 这步错了
应该是 =2sin(π/6+2x)+1

f（x）=2根号3sinxcosx+2cos²x-1
=根号3sin2x+cos2x
=2[sin2x*cosπ/6+cos2x*sinπ/6]
=2sin(2x+π/6)
(1)T=2π/2=π
最大值2和最小值-2
（2）2sin(2x0+π/6)=6/5
sin(2x0+π/6)=3/5
x0∈闭区间（π/4,π/2）
2x0+π/6∈闭区间（2π/3,7π/6）第二、三象限,cos(2x0+π/6)=-4/5
cos2x0=cos[(2x0+π/6)-π/6]
=cos(2x0+π/6)*cosπ/6+sin(2x0+π/6)*sinπ/6
-4/5*根号3/2+3/5*1/2
=(3-4根号3)/10

f(x)=2倍根号3*SinxCosx+2Cos²x 是这样吧=√3sin2x+cos2x+1=2sin(2x+π/6) +1(1) T=2π/2=π(2) 2kπ-π/2≤2x+π/6≤2kπ+π/22kπ-2π/3≤2x≤2kπ+π/3kπ-π/3≤x≤kπ+π/6增区间 【kπ-π/3,kπ+π/6】,k...

f(x)=2√3sinxcosx+2cos²x-1
=√3sin(2x)+cos(2x)
=2[(√3/2)sin(2x)+(1/2)cos(2x)]
=2sin(2x+π/6)
x1∈[π/4,π/2]
2π/3≤2x1+π/6≤7π/6 cos(2x1+π/6)

(1)
f(x)=2√3sinxcosx+2cos²x-1=√3sin2x+cos2x=2（√3/2sin2x+1/2cos2x）=2sin（2x+π/6）
函数的最小正周期T=π
(2)
f(x)=2sin（2x+π/6）=6/5
sin（2x+π/6）=3/5
x∈[π/4,π/2],
cos（2x+π/6）=-4/5
cos2x=cos(2x+π/6-π/6)=cos(2x+π/6)cosπ/6+sin(2x+π/6)sinπ/6=-4/5*√3/2+3/5*1/2=3/10-2√3/5

sin²x-sinxcosx+2cos²x
=1-sinxcosx+cos²x
=1-½sin2x+(1+cos2x)/2
=3/2+½cos2x-½sin2x
=3/2+(√2/2)cos(2x+π/4)
如果不懂,祝学习愉快!

f(x)=(1-cos2x)/2+(√3/2)sin2x+1+cos2x
f(x)=1/2+(cos2x)/2+(√3/2)sin2x+1
f(x)=3/2+sin(2x+π/6)
最值为+-5/2
周期2π/2=π
单增区间2kπ-π/2

f（x）=2根号3sinxcosx+2cos²x-1
=根号3sin2x+cos2x
=2[sin2x*cosπ/6+cos2x*sinπ/6]
=2sin(2x+π/6)
(1)T=2π/2=π
最大值2和最小值-2
（2）2sin(2x0+π/6)=6/5
sin(2x0+π/6)=3/5
x0∈闭区间（π/4,π/2）
2x0+π/6∈闭区间（2π/3,7π/6）第二、三象限,cos(2x0+π/6)=-4/5
cos2x0=cos[(2x0+π/6)-π/6]
=cos(2x0+π/6)*cosπ/6+sin(2x0+π/6)*sinπ/6
-4/5*根号3/2+3/5*1/2
=(3-4根号3)/10

f(x)=sin^2 x+√3sinxcosx+2cos^2 x,
=sin^2 x+cos^2 x+√3sinxcosx+cos^2 x,
=1+√3/2sin2x+1/2+1/2cos2x
=3/2+sin(2x+π/6)
最小正周期为：T=2π/2=π
单调增区间.
2kπ-π/2