复合函数f(y)=y²-y-2, y=cosx
cosx在[0+2kπ,π+2kπ]减,（π+2kπ,2π+2kπ]增
f(y)是典型的二次函数,y定义域[-1,+1],f(y)在[-1,-1/2]减,（-1/2,+1]增
cos(2π/3+2kπ)=cos(4π/3+2kπ)=-1/2,cos(π+2kπ)=-1,cos(0+2kπ)=1
∴x∈(2π/3+2kπ,4π/3+2kπ)时,y∈(-1,-1/2),f(x)减
x∈(－2π/3+2kπ,2π/3+2kπ)时,y∈(-1/2,+1),f(x)增
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画下图就明白,我也不知道数有没错,我没算,方法这样应该没错咯