f(x)=-√3sin^2x+sinxcosx求1.单调区间2在X∈[-π/2,0]的最最小值

Question

f(x)=-√3sin^2x+sinxcosx 求1.单调区间2在X∈[-π/2,0]的最最小值

Answer 1

原式=-√3(1-cos2x)/2+sin2x/2
=√3/2*cos2x+sin2x/2-√3/2
=sin60°cos2x+sin2xcos60°-√3/2
=sin(2x+π/3)-√3/2
sinx的单调递增区间为[2kπ -π/2,2kπ +π/2]
令2x+π/3∈[2kπ -π/2,2kπ +π/2]得
x∈[kπ -5π/12,kπ+π/12]
sinx的单调递减区间为（2kπ +π/2,2kπ +3π/2）
令2x+π/3∈（2kπ +π/2,2kπ +3π/2）得
x∈（kπ +π/12,kπ+7π/12）
故f（x）单调递增区间为[kπ -5π/12,kπ+π/12]
单调递减区间(kπ +π/12,kπ+7π/12）
(2)根据f（x）单调区间知道,
x∈[-π/2,-5π/12],f(x)单调递减,x∈（-5π/12,0]时候f(x)单调递增
所以f（x）在x=-5π/12时候取最小值
最小值为-1-√3/2
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