已知f(x)=sinx平方+2根号3sinxcosx-cosx平方 1)求f(x)的最大值和取最大值时候的集合 2)求f(x)的增区间
\u6025\uff01\u6025\uff01\u6025\uff01\u5df2\u77e5\u51fd\u6570f(x)=sin²x+2\u6839\u53f73sinxcosx-cos²x \uff081\uff09\u6c42f(x)\u53d6\u5f97\u6700\u5927\u503c\u65f6\u7684x\u503c\u3002f(x)=(1-cos2x)/2+\u221a3sin2x-(1+cos2x)/2=\u221a3sin2x-cos2x=2(sin2x*cos\u03c0/6-cos2x*sin\u03c0/6)=2(sin2x-\u03c0/6)
1)\u5f532x-\u03c0/6=2k\u03c0+\u03c0/2\u65f6\uff0cf(x)\u53d6\u6700\u5927\u503c\u4e3a2
\u6b64\u65f6x=k\u03c0+\u03c0/3, \u8fd9\u91cck\u4e3a\u4efb\u610f\u6574\u6570
2\uff09\u51fd\u6570\u7684\u5355\u8c03\u589e\u533a\u95f4\u4e3a\uff1a 2k\u03c0-\u03c0/2=<2x-\u03c0/6<=2k\u03c0+\u03c0/2
\u5373k\u03c0-\u03c0/6=<x<=k\u03c0+\u03c0/3
\u6240\u4ee5\u5728[0,\u03c0]\u5185\uff0c\u5355\u8c03\u589e\u533a\u95f4\u4e3a[0, \u03c0/3]U[5\u03c0/6, \u03c0]
f(x)=1+2sin²x+2\u221a3sinxcosx
=1+ 1-cos2x + \u221a3sin2x
=2+2(\u221a3/2sin2x - 1/2cos2x )
=2+2sin(2x-\u03c0/6)
f(x)\u7684\u6700\u5927\u503c=2+2\u00d71=4
\u5355\u8c03\u589e\u533a\u95f4\uff1a
\u4ee4-\u03c0/2 +2k\u03c0< 2x-\u03c0/6 <\u03c0/2+ 2k\u03c0
\u5f97 -\u03c0/6+k\u03c0 < x < \u03c0/3+k\u03c0
\u5373\u589e\u533a\u95f4\u662f\uff08-\u03c0/6+k\u03c0 \uff0c \u03c0/3+k\u03c0 \uff09\uff0ck\u662f\u6574\u6570\u3002
=√3sin2x-cos2x
=2sin(2x-π/6)
1)
最大值=2
此时2x-π/6=2kπ+π/2
x=kπ+π/3
2)
函数在区间满足：2kπ-π/2<=2x-π/6<=2kπ+π/2
即在区间：
【kπ-π/6，kπ+π/3】
单调递增。
f(x)=sinx平方+2根号3sinxcosx-cosx平方
= 2根号3sinxcosx-(cosx平方-sinx平方)
= 根号3sin2x - cos2x
= 2(sin2xcosπ/6 - cos2xsinπ/6)
= 2sin(2x-π/6)
最大值2
此时，2x-π/6=2kπ+π/2，x=kπ+π/3，其中k∈Z
2x-π/6∈（2kπ-π/2，2kπ+π/2）时单调增
增区间：（kπ-π/6，kπ+π/3）
f(x)=sin^2(x)+3^0.5*2sin(x)cos(x)-cos^2(x)
=3^0.5sin(2x)-cos(2x)
=2sin(2x+π/6)
最大值 2x+π/6=π/2+2kπ
x=π/6+kπ
最小值 2x+π/6=-π/2+2kπ
x=-π/3+kπ
f(x)的增区间
-π/2+2kπ≤2x+π/6≤π/2+2kπ
-π/3+kπ≤x≤π/6+kπ
f(x)=sin²x+2√3sinxcosx-cos²x
=√3sin2x-cos2x
=2(√3/2sin2x-1/2cos2x)
=2(cospi/6sin2x-sinpi/6cos2x)
=2sin(2x-pi/6)
f(x)最大值为2
2x-pi/6=pi/2+2kpi
x=pi/3+kpi
f(x)的增区间为
-pi/2+2kpi<2x-pi/6<pi/2+2kpi
-pi/6+kpi<x<pi/3+kpi
增区间(-pi/6+kpi,pi/3+kpi)
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