Question

求函数y=sinx(π/3-x/2),x属于[-2π,2π]的单调增区间

Answer 1

y=sinx(π/3-x/2)=-sinx(x/2-π/3),

x属于[-2π,2π],则x/2-π/3属于[-4π/3,2π/3],

所以x/2-π/3属于[-4π/3,-π/2],时,原函数递增,即[-2π,-π/3]