Question

已知函数f（x）=-√3 sin²x+sinxcosx 设α∈（0，π）f（α/2）=1/4-√3/2

求sina的 值

Answer 1

f(x)=-√3(1-cos2x)/2+1/2*sin2x
=1/2*sin2x+√3/2*cos2x-√3/2
=sin2xcosπ/3+cos2xsinπ/3-√3/2
=sin(2x+π/3)-√3/2

f(α/2)=sin(α+π/3)-√3/2=1/4-√3/2
sin(α+π/3)=1/4
α+π/3=arcsin(1/4)或α+π/3=π-arcsin1/4
α=arcsin(1/4)-π/3或α=2π/3-arcsin1/4

因为1/4<√3/2
所以arcsin(1/4)<π/3
而α>0
所以α=2π/3-arcsin1/4

Answer 2

x²+7x-12x-84=0
x(x+7)-12(x+7)=0
(x-12)(x+7)=0
x=12,x=-7