Find the value of $a$ such that $4\sin^2x+4\sin x+a^2-3<0$ for atleast one negative $x$ Find the value of $a$ such that $4\sin^2x+4\sin x+a^2-3<0$ for atleast one negative $x$.

$4\sin^2x+4\sin x+a^2-3<0$
$4\sin^2x+4\sin x<-a^2+3$
or $-a^2+3>4\sin^2x+4\sin x$
I found the minimum and maximum values of $4\sin^2x+4\sin x$.Minimum value is $-1$ and maximum value is $8$.
So $-a^2+3>8$
So $a^2+5<0$ But this gives me no values of $a$.And the answer given is $-\sqrt3<a<\sqrt3$.
I do not know where i am wrong.
 A: First of all, previous considerations:
a) For such $a$, if there is at least $x<0$ satisfiyng the inequality, then there is an $x^\prime\in[0,2\pi]$ that satisfies it (cause periodicity of sine function), and the inverse is also true. Thus, we can change the problem to find values of $a$ for which inequality is satisfied for at least one $x$ in $[0,2\pi]$.
b) Let $f:[0,2\pi]\to\mathbb{R}$ given by $f(x)=3-4\sin^2x-\sin x$. Then, it is easy to prove that $f(x)>0$ iff $x\in [0,\pi/6)\cup(5\pi/6,2\pi]$. 
Then, if $x\in[0,\pi/6)\cup(5\pi/6,2\pi]$, there is an $a$ such that $f(x)>a^2$, due to (b). But, in $[0,\pi/6)\cup(5\pi/6,2\pi]$, the maximum are attained at $x=0,7\pi/6,11\pi/6$, yielding $f(x)=3,4,4$ respectively.
Putting all together: By means of Intermediate Value Theorem, for each $y\in(0,4]$, there is an $x\in[0,\pi/6)\cup(5\pi/6,2\pi]$ such that $f(x)=y$. Finally, if we take $a^2\in[0,4)$, there are $y\in(a^2,4)$ and $x\in[0,\pi/6)\cup(5\pi/6,2\pi]$ such that $f(x)=y>a^2$. So, the interval for $a$ is when $a^2\in[0,4)$, which is equivalent to $-2<a<2$.
Conversely, by the same idea, it is easy to show that for $a\in(-2,2)$ there is an $x\in[0,\pi/6)\cup(5\pi/6,2\pi]$ such that $f(x)>a$.
Conclusion: the desired interval is (-2,2), not $(-\sqrt{3},\sqrt{3})$ 
