已知集合A={(x,y)|{(x≥1),(y≤1),(x-y≤根号2)},集合B={(x,y)|xcosα+ysinα-1=0已知集合A={(x,y)|{(x≥1),(y≤1),(x-y≤2)},集合B={(x,y)|xcosα+ysinα-1=0,α∈[0,2π)},若A交B≠∅,则α的取值范围?
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/27 03:05:55
![已知集合A={(x,y)|{(x≥1),(y≤1),(x-y≤根号2)},集合B={(x,y)|xcosα+ysinα-1=0已知集合A={(x,y)|{(x≥1),(y≤1),(x-y≤2)},集合B={(x,y)|xcosα+ysinα-1=0,α∈[0,2π)},若A交B≠∅,则α的取值范围?](/uploads/image/z/8574209-17-9.jpg?t=%E5%B7%B2%E7%9F%A5%E9%9B%86%E5%90%88A%3D%7B%28x%2Cy%29%7C%7B%28x%E2%89%A51%29%2C%28y%E2%89%A41%29%2C%28x-y%E2%89%A4%E6%A0%B9%E5%8F%B72%29%7D%2C%E9%9B%86%E5%90%88B%3D%7B%28x%2Cy%29%7Cxcos%CE%B1%2Bysin%CE%B1-1%3D0%E5%B7%B2%E7%9F%A5%E9%9B%86%E5%90%88A%3D%7B%EF%BC%88x%EF%BC%8Cy%EF%BC%89%7C%7B%28x%E2%89%A51%29%2C%28y%E2%89%A41%29%EF%BC%8C%28x-y%E2%89%A42%29%7D%2C%E9%9B%86%E5%90%88B%3D%7B%28x%2Cy%29%7Cxcos%CE%B1%2Bysin%CE%B1-1%3D0%2C%CE%B1%E2%88%88%5B0%2C2%CF%80%29%7D%2C%E8%8B%A5A%E4%BA%A4B%E2%89%A0%26%238709%3B%2C%E5%88%99%CE%B1%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%EF%BC%9F)
已知集合A={(x,y)|{(x≥1),(y≤1),(x-y≤根号2)},集合B={(x,y)|xcosα+ysinα-1=0已知集合A={(x,y)|{(x≥1),(y≤1),(x-y≤2)},集合B={(x,y)|xcosα+ysinα-1=0,α∈[0,2π)},若A交B≠∅,则α的取值范围?
已知集合A={(x,y)|{(x≥1),(y≤1),(x-y≤根号2)},集合B={(x,y)|xcosα+ysinα-1=0
已知集合A={(x,y)|{(x≥1),(y≤1),(x-y≤2)},集合B={(x,y)|xcosα+ysinα-1=0,α∈[0,2π)},若A交B≠∅,则α的取值范围?
已知集合A={(x,y)|{(x≥1),(y≤1),(x-y≤根号2)},集合B={(x,y)|xcosα+ysinα-1=0已知集合A={(x,y)|{(x≥1),(y≤1),(x-y≤2)},集合B={(x,y)|xcosα+ysinα-1=0,α∈[0,2π)},若A交B≠∅,则α的取值范围?
[Answer]
本来是 -π/2