y=sin^2x+2sinx*cosx+3cos^2x
=sin2x+cos2x+2
=根号2sin(2x+pi/4)+2
所以最小值是2－根号2
次时
2x+pi/4=2kpi-pi/2
x=kpi-3pi/8
y=sin^2x+2sinx*cosx+3cos^2x
1个回答
相关问题
-
y=sinx/(2-cosx) =2sinx*cosx/[3sin^2(x/2)+cos^(x/2)] =2/[3tan
-
求cos(4sinx)、cos(3sinx)、sin(cosx)、sin(cos2x)、sin(cos3x)、sin(2
-
sinx+cosx/sinx-cosx=2 求sinx/cos^3x +cosx/sin^3x
-
SinX+cosX=3+2根号2,求sin2X+2cos2x分之sin2x+根号2sinx*cosx-cos2x
-
求证:(1-2sinx×cosx)/cos2x-sin2x=(cos2x-sin2x)/(1+2sinx×cosx)
-
f(x)=2cos*sin(x+π/3)-^3sin^2x+sinx*cosx
-
证明cos3x/sinx+sin3x/cosx=2cot2x
-
2sin(x-π/4）sin（x+π/4）=（sinx-cosx）（sinx+cosx）=-cos2a
-
2sin^2*x-sinx*cosx-cos^2*x=1
-
若sinx+cosx=1/5,求cosx,sinx,sin2x,cos2x