Question

已知f想)=cosx(cosx-3)+sinx(sinx-3)

1.若x属于(2π.3π)求函数f(x)的单调增区间.
（2）若x属于（π/2.3π/4）且f(x)=-1.求tan2x.

Answer 1

f(x)=cosx(cosx-3)+sinx(sinx-3)
=(cosx)^2-3cosx+(sinx)^2-3sinx
=1-3(sinx+cosx)
=1-3√2sin(x+π/4)

(1)
求f(x)的单调增区间
就是求y=sin(x+π/4)的单调减区间
令2kπ+π/2＜x+π/4＜2kπ+3π/2(k∈Z)
2kπ+π/4＜x＜2kπ+5π/4(k∈Z)

因为x∈(2π,3π)
所以9π/4＜x＜3π
即函数f(x)的单调增区间是(9π/4,3π)
(2)
f(x)=-1
所以1-3√2sin(x+π/4)=-1
故sin(x+π/4)=2/3√2=√2/3

所以cos(2x+π/2)=1-2[sin(x+π/4)]^2=1-2*(√2/3)^2=5/9
所以sin2x=-5/9

x∈(π/2,3π/4)
2x∈(π,3π/2)
所以cos2x=-√[1-(-5/9)^2]=-2√14/9

所以tan2x=sin2x/cos2x=(-5/9)/(-2√14/9)=5√14/28

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Answer 2

已知f(x)=cosx(cosx-3)+sinx(sinx-3)，1.若x属于(2π.3π)求函数f(x)的单调增区间；
（2）若x属于（π/2.3π/4）且f(x)=-1.求tan2x.
解：
(1)f(x)=cos²x-3cosx+sin²x-3sinx=1-3(sinx+cosx)=1-3(√2)sin(x+π/4)，
在(2π.3π)内的单增区间为：[9π/4，3π)。
(2).f(x)=1-3(sinx+cosx)=-1，故sinx+cosx=2/3，1+sin2x=4/9，sin2x=4/9-1=-5/9，
π/2<x<3π/4，π<2x<3π//2，cos2x=-√(1-25/81)=-√(56/81)=-(√56)/9，
故tan2x=sin2x/cos2x=5/√56=5/(2√14)=(5/28)√14.

Answer 3

f(x)=cos²x-3cosx+sin²x-3sinx
=1-3√2(√2/2cosx+√2/2sinx)
=1-3√2sin(x+π/4)
(1)令2kπ+π/2≤x+π/4≤2kπ+3π/2 k∈N
2kπ+π/4≤x≤2kπ+5π/4 k∈N
单调增区间为[2kπ+π/4,2kπ+5π/4] (k∈N)
(2)f(x)=1-3√2sin(x+π/4)=-1
sin(x+π/4)=√2/3
x∈(π/2,3π/4) 那么3π/4≤x+π/4≤π
cos(x+π/4)=-√7/3
tan(x+π/4)=-√2/√7
tan2x=-cot(2x+π/2)=-1/tan2(x+π/4)=[tan²(x+π/4)-1]/2tan(x+π/4)=5√14/28