f(x)=4sin(x-派/6)cosx+1怎么求单调区间?

f(x)=4sin(x-派/6)cosx+1怎么求单调区间?
f(x)=4sin(x-派/6)cosx+1
怎么求单调区间?

f(x)=4sin(x-派/6)cosx+1怎么求单调区间?
f(x)=4sin(x-π/6)cosx+1
=4[sinxcos(π/6)-cosxsin(π/6)]cosx+1
=4[(√3/2)sinx -(1/2)cosx]cosx +1
=2√3sinxcosx -2cos²x +1
=√3(2sinxcosx)-(2cos²x-1)
=√3sin(2x)-cos(2x)
=2[(√3/2)sin(2x)-(1/2)cos(2x)]
=2sin(2x-π/6)
剩下的就简单了.
2kπ-π/2≤2x-π/6≤2kπ+π/2 (k∈Z)时,f(x)单调递增,此时kπ-π/6≤x≤kπ+π/3 (k∈Z)
2kπ+π/2≤2x-π/6≤2kπ+3π/2 (k∈Z)时,f(x)单调递减,此时kπ+π/3≤x≤kπ+5π/6 (k∈Z)
函数的单调递增区间为[kπ-π/6,kπ+π/3],单调递减区间为[kπ+π/3,kπ+5π/6] (k∈Z)

答：
f(x)=4sin(x-π/6)cosx+1
=2*[sin(x-π/6+x)+sin(x-π/6-x)]+1
=2sin(2x-π/6)-2*sin(π/6)+1
=2sin(2x-π/6)
单调递增区间满足：
2kπ-π/2<=2x-π/6<=2kπ+π/2，kπ-π/6<=x<=kπ+π/3
单调递减区间满足：
2kπ+π...

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答：
f(x)=4sin(x-π/6)cosx+1
=2*[sin(x-π/6+x)+sin(x-π/6-x)]+1
=2sin(2x-π/6)-2*sin(π/6)+1
=2sin(2x-π/6)
单调递增区间满足：
2kπ-π/2<=2x-π/6<=2kπ+π/2，kπ-π/6<=x<=kπ+π/3
单调递减区间满足：
2kπ+π/2<=2x-π/6<=2kπ+3π/2，kπ+π/3<=x<=kπ+5π/6
所以：
单调递增区间[kπ-π/6，kπ+π/3]
单调递减区间[kπ+π/3，kπ+5π/6]
以上k属于Z

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