解一下这道题,高一数学的有关(三角函数)

2025-05-09 23:21:49
推荐回答(2个)
回答1:

√2sin(x+π/4)=2m
sin(x+π/4)=√2m
0到π之间(闭区间)
π/4<=x+π/4<=5π/4
-√2/2<=sin(x+π/4)<=1,即-√2/2<=√2m<=1
所以,-1/2<=m<=√2/2
画一个sin的图(注意取值范围,π/4<=x+π/4<=5π/4)
所以要有相异两实根
只能是π/4<=x+π/4<=3π/4时
√2/2<=√2m<1
1/2<=m<√2/2
所以,1/2<=m<√2/2

回答2:

方程式左方根号下么