y=3sin(π 3-x 2)单调增区间

最大值y=2*1+3=5最小值y=2*（-1）+3=1令2kπ-π/2≤x-π/3≤2kπ+π/2得2kπ-π/6≤x≤2kπ+5π/6,k为整数令2kπ+π/2≤x-π/3≤2kπ+3π/2得2kπ

y=-sin(x/2-π/3)所以y增区间即sin(x/2-π/3)的减区间sinx的减区间是(2kπ+π/2,2kπ+3π/2)2kπ+π/2<x/2-π/3<2kπ+3π/22kπ+5

π/2+2kπ再问：换元法有没有？再答：令3x+π/4=t，y=2sint的递减区间是：π/2+2kπ

y=-1/2sin(2/3x-π/4)所以y和sin(2/3x-π/4)单调性相反sinx的增区间是(2kπ-π/2,2kπ+π/2)减区间是(2kπ+π/2,2kπ+3π/2)所以sin(2/3x-

y=sin(π/3-2x)=-sin(2x-π/3)2kπ+π/2≤2x-π/3≤2kπ+3π/2kπ+5π/12≤x≤kπ+11π/12故增区间是[kπ+5π/12,kπ+11π/12]

化为y=-sin(3x/2-π/4),由题,π/2+2kπ≤3x/2-π/4≤3π/2+2kπ,3π/4+2kπ≤3x/2≤7π/4+2kπ,π/2+4kπ/3≤x≤7π/6+4kπ/3,k属于Z递增

y=sin（x2+π3）的单调递增区间由2kπ-π2≤x2+π3≤2kπ+π2（k∈Z）得：4kπ-5π3≤x≤4kπ+π3（k∈Z），∵x∈[-2π，2π]，∴-5π3≤x≤π3．即y=sin（x2

步骤二当2kπ+π/2≤m≤2kπ+3π/2时,y=3sinm随x增大而减小.第二步错了,应该改为“当2kπ+π/2≤m≤2kπ+3π/2时,y=3sinm随m增大而减小"正确的解答过程：法一：步骤一

y=3sin(3x+π/4)单调增区间是:2kPai-Pai/2

把两个三角函数展开,得y=3/2sinx-√3/2cosx合并成：y=√3sin(x-π/6)单调区间是（-π/3,2π/3）增（2π/3,5π/3）减其中都要加上2kπ,我就不写了

y=3sin(π/3-2x)=-3sin(2x-π/3)在2x-π/3∈(2kπ+π/2,2kπ+3π/2)时单调增故单调增区间是：x∈(kπ+5π/12,kπ+11π/12)k∈Z如仍有疑惑,欢迎追

你的答案没有错啊增区间是【-7/3π-4kπ,-π/3-4kπ】,k∈Z只不过习惯写成增区间是【-7/3π+4kπ,-π/3+4kπ】,k∈Z再问：答案是[5π/3+4kπ，11π/3+4kπ](k∈

y=sin(2x+π/3)+cos(2x-π/6)=(1/2)sin2x+(√3/2)cos2x+(√3/2)cos2x+(1/2)sin2x=sin2x+√3cos2x=2sin(2x+π/3)2k

y=sinx的单调递增区间是[2kπ-π/2,2kπ+π/2](k∈Z)y=sin(π/3-x/2)是由函数t=π/3-x/2与函数y=sint复合而成.t=π/3-x/2是一直递减的,函数y=sin

(-5π/18+2kπ,π/18+2kπ)单增（π/18+2kπ,7π/18+2kπ）单减Ymax=2Ymin=-2中心对称轴kπ/3-π/9

sinx单调递减区间是（π/2+2kπ,3π/2+2kπ）.所以,当2x+π/3∈（π/2+2kπ,3π/2+2kπ）时,函数单调递减x∈（π/12+kπ,7π/12+kπ）.（k=0,1,2,3……

－π/2+2kπ《π/3-x/2《π/2+2kπ得单调递增区间为[-π/12+kπ,5π/12+2kπ]

解由y=3sin(π/3-x/2)=-3sin(x/2-π/3)故当2kπ+π/2≤x/2-π/3≤2kπ+3π/2,k属于Z时,y是增函数,故当4kπ+5π/3≤x≤4kπ+11π/3,k属于Z时,

y=-3sin(2x-π/3)-π/2+2kπ＜2x-π/3＜π/2+2kπ-π/12+kπ＜x＜5π/12+kπ单调减区间（-π/12+kπ,5π/12+kπ）k∈Zπ/2+2kπ＜2x-π/3＜3

y=sinx增区间[2kπ-π/2,2kπ+π/2]所以本题,2kπ-π/2≤π/4+2x≤2kπ+π/2kπ-3π/8