求函数y=2sin(2x+π/6)+2sin(2x-π/12)的最大值,并求出此时x的集合.
问题描述:
求函数y=2sin(2x+π/6)+2sin(2x-π/12)的最大值,并求出此时x的集合.
答
令t=2x-π/12,则2x+π/6=2t+π/4,所以y=2sin(2x+π/6)+2sin(2x-π/12)=2sin(t+π/4)+2sint=√2sint+√2cost+2sint=(√2+2)sint+√2cost=2√(2+√2)sin(t+θ)(其中tanθ=√2-1)所以原函数的最大值为2√(2+...