Question

（1）求y=2sin2x+2sinx-1的值域 （2）求y=3-2cos(2x+π/3),x∈[-π/3,π/6]的最值

Answer 1

y=2sin²x+2sinx-1

= 2(sinx+1/2)^2-3/2

-1≤sinx≤1

-1/2≤sinx+1/2≤3/2

0≤(sinx+1/2)²≤9/4

-3/2≤(sinx+1/2)²≤3/4

值域【-3/2,3/4】

y=3-2cos(2x+π/3)

x∈[-π/3,π/6]

2x∈[-2π/3,π/3]

2x+π/3∈[-π/3,π]

2x+π/3=0时,有最小值3-2=1

2x+π/3=π时,有最大值3+2=5