How to solve this inequality for $a$?
$\cos^2x + (a+1)\cos x+a<0 : x \in R$
This what i have tried so far.
By factoring, $(\cos x+1)(\cos x+a)<0$
Since, $(\cos x+1)\in [0,2] , (\cos x+a)<0$
After this I can't proceed any further. Is my approach any good in this case? Or is there a better alternative for this?
Right so go on:
$(\cos x + 1)(\cos x + a) < 0$ so either:
1) $\cos x + 1 > 0$ and $\cos x + a < 0$ or
2) $\cos x + 1 < 0$ and $\cos x + a > 0$.
1) can be true if $x \ne (2k+1) \pi$ and if $a < - \cos x$. But this will not always be true. If $a \le -1$ the will only not be true with $x = (2k+1)\pi$.
2) can never be true
However as $\cos x + 1 = 0$ is a possibility so there is no $a$ where this is always true. So there is not solution.