已知向量a=(sinx,2√ 3sinx),b=(mcosx,-sinx),定义f(x)=a*b+√ 3,且x=π/6是函数Y=F(X)的零点(1)求函数y=f(x)在R上的单点区间2.若函数y=f(x+θ)(0
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![已知向量a=(sinx,2√ 3sinx),b=(mcosx,-sinx),定义f(x)=a*b+√ 3,且x=π/6是函数Y=F(X)的零点(1)求函数y=f(x)在R上的单点区间2.若函数y=f(x+θ)(0](/uploads/image/z/3807267-51-7.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8Fa%3D%28sinx%2C2%E2%88%9A+3sinx%29%2Cb%3D%28mcosx%2C-sinx%29%2C%E5%AE%9A%E4%B9%89f%28x%29%3Da%2Ab%2B%E2%88%9A+3%2C%E4%B8%94x%3D%CF%80%2F6%E6%98%AF%E5%87%BD%E6%95%B0Y%3DF%28X%29%E7%9A%84%E9%9B%B6%E7%82%B9%EF%BC%881%EF%BC%89%E6%B1%82%E5%87%BD%E6%95%B0y%3Df%28x%29%E5%9C%A8R%E4%B8%8A%E7%9A%84%E5%8D%95%E7%82%B9%E5%8C%BA%E9%97%B42.%E8%8B%A5%E5%87%BD%E6%95%B0y%3Df%28x%2B%CE%B8%29%280)
已知向量a=(sinx,2√ 3sinx),b=(mcosx,-sinx),定义f(x)=a*b+√ 3,且x=π/6是函数Y=F(X)的零点(1)求函数y=f(x)在R上的单点区间2.若函数y=f(x+θ)(0
已知向量a=(sinx,2√ 3sinx),b=(mcosx,-sinx),定义f(x)=a*b+√ 3,且x=π/6是函数Y=F(X)的零点
(1)求函数y=f(x)在R上的单点区间
2.若函数y=f(x+θ)(0<θ<π/2)为奇函数,求θ
3.在三角形ABC中,a,b,c分别是角A,B,C的对边,已知a=1,b=√2,f(A)=-1,求角C
已知向量a=(sinx,2√ 3sinx),b=(mcosx,-sinx),定义f(x)=a*b+√ 3,且x=π/6是函数Y=F(X)的零点(1)求函数y=f(x)在R上的单点区间2.若函数y=f(x+θ)(0
(1)
f(x)=msinxcosx-2√3sin^2(x)+√3 ,因为x=π/6是函数的零点,所以
0=m(√3/4)-√3/2+√3 ==>m= - 2
f(x)= - sin2x-√3(1-cos2x)+√3
=2cos(2x+π/6)
由 -π+2kπ≤2x+π/6≤2kπ得单调增区间是:【-7π/12+kπ,-π/12+kπ】
由 2kπ≤2x+π/6≤π+2kπ得单调增区间是:【-π/12+kπ,5π/12+kπ】
(2)
f(x+θ)=2cos(2x+2θ+π/6)是奇函数,所以当x=0时,上式为零
0=2cos(2θ+π/6)=0 ==>θ=π/6
(3)
2cos(2A+π/6)= - 1 ==>cos(2A+π/6)= - 1/2
A=π/4 ,则正弦定理得:1/sinπ/4=√2/sinB ==>sinB=1
B=π/2 ==>C=π/4