我想是可以的
f'(x)=[(sinx)'(2+cosx)-(2+cosx)'sinx]/[(2+cosx)^2]
=[2cosx+(cosx)^2+(sinx)^2]/[(2+cosx)^2]
=[2cosx+1]/[(2+cosx)^2]
令t=2+cosx,1<=t<=3
則f'(x)=[2t-3]/[t^2]
令f'(x)=0,解得t=3/2,即cosx=-1/2,x=2kπ+2π/3(k屬于整數(shù))
當(dāng)t屬于[1,3/2)即cosx屬于[-1,-1/2)時(shí)f'(x)<0（代-1進(jìn)f'(x)算）
當(dāng)t屬于(3/2,3]即cosx屬于(-1/2,1]時(shí)f'(x)>0
**最后的區(qū)間怎么取我不是很確定
f(x)單調(diào)遞減區(qū)間[2kπ+2π/3,(2k+1)π)
f(x)單調(diào)遞減區(qū)間[2kπ,2kπ+2π/3)
最后祝新年快樂(lè)