Question

求函数y=2sin(2x+π/3),x属于(-π/6,π/6)的值域？

Answer 1

x∈(-π/6,π/6)
2x∈(-π/3,π/3)
2x+π/3∈(0,2π/3)
sin(2x+π/3)∈(0,1]
2sin(2x+π/3)∈(0,2]
所以函数y=2sin(2x+π/3),x属于(-π/6,π/6)的值域为：(0,2],8,
越南风情 举报
为什么sin2π/3=1? 2x+π/3∈(0,2π/3) 当2x+π/3=π/2时 sin(2x+π/3)=1,x属于(-π/6,π/6)
2x+π/3属于(0,2π/3)
值域(0,2],0,x属于(-π/6,π/6)
(2x+π/3)属于(0,2π/3)
2sin(2x+π/3)属于(0,1],0,求函数y=2sin(2x+π/3),x属于(-π/6,π/6)的值域
要过程