f(x)=cos^2x+cosxsinx(0<x<π/2) 1.若f(x)=1,求x的值 2,求f(x)的值域f(x)=cos^2x+cosxsinx(0<x<π/2)1.若f(x)=1,求x的值2,求f(x)的值域

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/25 17:17:50
f(x)=cos^2x+cosxsinx(0<x<π/2) 1.若f(x)=1,求x的值 2,求f(x)的值域f(x)=cos^2x+cosxsinx(0<x<π/2)1.若f(x)=1,求x的值2,求f(x)的值域
x)KӨдM/3Rřy̩ F z/aiMR>UF`e Ka1XfjiŌ*lu25R@L4Ar`CM\OA:-ek "t*lA4̅gFBtLA0cm @A $x

f(x)=cos^2x+cosxsinx(0<x<π/2) 1.若f(x)=1,求x的值 2,求f(x)的值域f(x)=cos^2x+cosxsinx(0<x<π/2)1.若f(x)=1,求x的值2,求f(x)的值域
f(x)=cos^2x+cosxsinx(0<x<π/2) 1.若f(x)=1,求x的值 2,求f(x)的值域
f(x)=cos^2x+cosxsinx(0<x<π/2)
1.若f(x)=1,求x的值
2,求f(x)的值域

f(x)=cos^2x+cosxsinx(0<x<π/2) 1.若f(x)=1,求x的值 2,求f(x)的值域f(x)=cos^2x+cosxsinx(0<x<π/2)1.若f(x)=1,求x的值2,求f(x)的值域
f(x)=cos^2x+cosxsinx=(cos2x+1)/2+1/2sin2x=√2/2sin(2x+π/4)+1/2
①f(x)=1 ,1=√2/2sin(2x+π/4)+1/2 ,√2/2=sin(2x+π/4) ,2x+π/4=3π/4 ,x=π/4
②值域[-√2/2+1/2,√2/2+1/2]