Question

f（x）=2根号3sinxcosx+2cos^2 x-1（x属于R）求增区间及[0，兀/2]的值域.若f（x0）=6/5，x0属于[兀/4，兀

f（x）=2根号3sinxcosx+2cos^2 x-1（x属于R）求增区间及[0，兀/2]的值域.若f（x0）=6/5，x0属于[兀/4，兀/2]，求cos2x0的值.

Answer 1

f（x）=2√3sinxcosx+2cos^2 x-1
=√3sin2x+cos2x......正弦余弦二倍角公式
=2sin(2x+π/6).....辅助角公式
令-π/2+2kπ<=2x+π/6<=π/2+2kπ,k∈Z
∴-π/3+kπ<=x<=π/6+kπ,k∈Z
∴增区间是[-π/3+kπ,π/6+kπ],k∈Z
x∈[0,π/2]
2x+π/6∈[π/6,7π/6]
2sin(2x+π/6)∈[-1,2]
值域是[-1,2]
(2)
x0∈[π/4，π/2]
2x0∈[π/2，π],第二象限
f(x0)=2sin(2x0+π/6)=6/5
sin(2x0+π/6)=3/5
cos(2x0+π/6)=-4/5
cos(2x0)
=cos(2x0+π/6-π/6)
=cos(2x0+π/6)cosπ/6+sin(2x0+π/6)sinπ/6
=-4/5*√3/2+3/5*1/2
=(3-4√3)/10
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