问题
选择题
设x∈（0，π），关于x的方程2Sin(x+
|
答案
∵x∈（0，π），∴
＜x+π 3
＜π 3
，∴-4π 3
＜sin（x+3 2
）≤1，π 3
由于关于x的方程2Sin(x+
)=a有2个不同的实数解，π 3
∴
＜3 2
＜1，∴a 2
＜a＜2，3
故选C．
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设x∈（0，π），关于x的方程2Sin(x+
|
∵x∈（0，π），∴
＜x+π 3
＜π 3
，∴-4π 3
＜sin（x+3 2
）≤1，π 3
由于关于x的方程2Sin(x+
)=a有2个不同的实数解，π 3
∴
＜3 2
＜1，∴a 2
＜a＜2，3
故选C．
| 任务阅读。阅读短文，根据短文内容完成表格。 | ||||||||||||||
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