y=sin(2x+π/6),x∈[π/4,3π/4]求值域
2个回答
展开全部
π/6<=x<=3π/4
π/3<=2x<=3π/2
π/2<=2x+π/6<=5π/3
π/2<=x<=3π/2
sinx是减函数
所以sin(3π/2)<=sin(2x+π/6)<=sin(π/2)
-1<=sin(2x+π/6)<=1
而sin(2x+π/6)值域本身就是[-1,1]
所以3π/2<=2x+π/6<=5π/3肯定也属于[-1,1]
π/3<=2x<=3π/2
π/2<=2x+π/6<=5π/3
π/2<=x<=3π/2
sinx是减函数
所以sin(3π/2)<=sin(2x+π/6)<=sin(π/2)
-1<=sin(2x+π/6)<=1
而sin(2x+π/6)值域本身就是[-1,1]
所以3π/2<=2x+π/6<=5π/3肯定也属于[-1,1]
追问
答案不是这个
追答
看错了,对不起
π/4<=x<=3π/4
π/2<=2x<=3π/2
2π/3<=2x+π/6<=5π/3
-1=sin(3π/2)<= sin(2x+π/6)=<sin(2π/3)=√3/2
所以
-1<=sin(2x+π/6)<=√3/2
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