求解一道高中数学函数题 已知f(x)=cosx(√3sinx+cosx) 求当x属于[0，π⼀2],求函数的最大值及取最大值时的x

f(x)=[3]cosxsinx+cosx^2=[3]/2sin2x+1/2(cos2x+1)=[3]/2sin2x+1/2cos2x+1/2
=cos30osin2x+sin30ocos2x+1/2
=sin(2x+π/6)+1/2
0

f(x)=cosx(√3sinx+cosx)
=√3sinxcosx+cos^2 x
=√3/2 sin2x+(1+cos2x)/2
=sin(2x+π/6)+1/2
0<=x<=π/2
0<=2x<=π
π/6<=2x+π/6<=7π/6
-1/2<=sin(2x+π/6)<=1
2x+π/6=π/2
x=π/6
y max=3/2

f(x)=cosx(√3sinx+cosx) =√3sinxcosx+cos^2 x
=√3/2 sin2x+(1+cos2x)/2=sin(2x+π/6)+1/2
0<=x<=π/2 0<=2x<=π
π/6<=2x+π/6<=7π/6 -1/2<=sin(2x+π/6)<=1
当2x+π/6=π/2，即x=π/6 函数最大值=3/2
还不懂的话联系我

最大值(根3+根2)/2
最小值(根2-1)/2

肯定先化简！里面提起来！ 讨论就行了！