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[函数] $2\cos^2x-3a\sin x+a-1=0$有三个不等实根,求$a$

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abababa posted 2015-4-15 20:36 |Read mode
$2\cos^2x-3a\sin x+a-1=0$在$[0,2\pi]$有三个不等实根,求$a$
$a=-1,-\frac{1}{2},\frac{1}{4}$时有三个不等实根,怎么证明其它情况没有三个不等实根呢?

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luren8asdf posted 2015-4-17 16:18
这是一个关于sinx的二次函数,如果二次函数有两个不同的解,则其他一个是1或-1,另一个在-1和1之间;如果二次函数有等根,则这个跟只能是0.
正好对于上面的三个a的值。

不会使用代码

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original poster abababa posted 2015-4-18 16:05
回复 2# luren8asdf
我没看懂,方程是$2\sin^2x-3a\sin x-(a+1)=0$,我用软件画图时确实是上面这三个解,其它情况就会有四个或者两个实根。为什么恰有三个实根时$\sin x$只能取$0,\pm1$?怎么证明它呢?能再详细讲讲吗?

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