Question

已知函数F(X)=2倍根号3sinx/2*cosx/2-(cos^2x/2-sin^2x/2).求fx最大值并求出此时x？

Answer 1

F(X)=2√3sinx/2*cosx/2-(cos^2x/2-sin^2x/2).
=√3sinx-cosx
=2(√3/2sinx-1/2cosx)
=2sin(x-π/6)
当x-π/6=2kπ+π/2,
即x=2kπ+2π/3,k∈Z时
f(x)取得最大值2
若f(x)=0,则sin(x-π/6)=0
∴x-π/6=kπ
∴x=k+π/6,k∈Z
∴k为偶数时,sinx=1/2,cosx=√3/2
sinx+cos[π+x]/sinx+sin[π/2-x]
=sinx-cosx/sinx+cosx
=1/2-√3+√3/2
=(1+√3)/2
k为奇数时,sinx=-1/2,cosx=-√3/2
原式=sinx-cosx/sinx+cosx
=-1/2-√3-√3/2
=-(1+3√3)/2,6,已知函数F(X)=2倍根号3sinx/2*cosx/2-(cos^2x/2-sin^2x/2).求fx最大值并求出此时x的值
若fx=0 ,求sinx+cox[π+x]/sinx+sin[π/2-x]的值