Question

数学题第21题，第三小题不会，求手写过程

Answer 1

f(0)=b+2=3, b=1
f(x)=1/2sin2x+(cos2x+1)/2+2=1/2sin2x+1/2cos2x+5/2=根号2/2sin(2x+Pai/4)+5/2
f(x)=5/2,即有sin(2x+Pai/4)=0
即有sin2x+cos2x=0
tan2x=-1.
tan2x=2tanx/(1-tan^2x)=-1
2tanx=tan^2x-1
由于tanx>0
故有:tanx=(2+2根号2)/2=根号2+1
tan(x+Pai/4)=(tanx+1)/(1-tanx)=(根号2+2)/(-根号2)=-根号2-1

Answer 2

F(0)=b+2=3
b=1
f(x)=sinxcosx+cos^2x+2
=1/2sin2x+1/2(1+cos2x)+2
=√2/2sin（2x+π/4)+5/2
f(x)最大值＝（5+√2)/2
f(x)最小值＝（－5+√2)/2
2）f(x)=√2/2sin（2x+π/4)+5/2=5/2
√2/2sin（2x+π/4)=0
sin2x+cos2x=0
tan2x= -1
tan2x= 2tanx/(1-tan^2x)=-1
tan^2x-2tanx-1=0
tanx=1+√2
tan(x+π/4)=(1+tanx)/(1-tanx)=-(2+√2)/√2=-(2+2√2)/2