1.求函数y=1+sinx+cos(x-π/6)的单调区间和最值,并指出取得最值时x的取值

1、y = 1+sinx + √3 /2 cosx + 1/2 sinx
= 3/2 sinx + √3/2 cosx + 1
= √3 sin (x+ π/6) + 1
-π/2 < x+π/6 < π/2 增区间 -2π/3 < x < π/3
π/2 < x+π/6 < 3π/2 减区间 π/3 < x < 4π/3
当 x+π/6 = π/2 +2kπ 即 x = π/3 + 2kπ (k∈Z)时
取最大值 √3 +1
2、x=π/8 是对称轴,则 x=π/8时取最值
2*π/8 +α = π/2 + kπ
α = π/4 + kπ
∵α∈（-π,0）
∴α = - 3π/4
- π/2 < 2x - 3π/4 < π/2
增区间 π/8 < x < 5π/8