Range of $a$ in Trigonometric equation 
If the inequality $\sin^2 x+a\cos x+a^2>1+\cos x$ hold for any $x\in \mathbb{R}.$ Then range of $a$ is 

Try: $1-\cos^2 x+a\cos x+a^2-1-\cos x>0$
$$\cos^2 x+(1-a)\cos x-a^2<0$$
$$4\cos^2 x+4(1-a)\cos x-4a^2<0$$
$$\bigg(2\cos x+(1-a)\bigg)^2-4a^2-(1-a)^2<0$$
Could some help me how to solve further, Thanks

Alternatively, write it as:
$$\sin^2 x+a\cos x+a^2>1+\cos x \iff \\
a^2+\cos x\cdot a-(\cos ^2x+\cos x)>0 \iff \\
\left(a+\frac12\cos x\right)^2-(\cos ^2x +\cos x+\frac 14\cos ^2x)>0$$
Note that the graph of $f(a)=\left(a+\frac12\cos x\right)^2-(\cos ^2x +\cos x+\frac 14\cos ^2x)$ is an opening-up parabola with the vertex:
$$V\left(-\frac12\cos x,-\cos ^2x-\cos x-\frac14\cos^2x\right)$$
The extreme $y$-coordinates of the vertex are:
$$f'(x)=\left(-\cos^2x -\cos x-\frac14\cos^2x\right)'=\frac52\cos x\sin x+\sin x=0 \Rightarrow \\
1) \ \sin x\left(\frac52\cos x+1\right)=0 \Rightarrow 
\sin x=0 \ \text{(which implies $\cos x=\pm 1)$}\ \text{or} \\ 
2) \ \cos x=-\frac25 \ \text{(which imples $\sin x=\pm \frac{\sqrt{21}}{5}$)}.$$
Hence, there are three critical points:
$$\cos x=-1 \ \left(\text{$V\left(\frac12,-\frac14\right)$ is local min}\right);\\
\cos x=-\frac25 \ \left(\text{$V\left(\frac15,\frac15\right)$ is local max}\right);\\
\cos x=1 \ \left(\text{$V\left(-\frac12,-\frac94\right)$ is local min}\right)$$
Note: 1) SOC are left as an exercise. 2) See the Desmos graph.  
Hence, it is sufficient to consider the extreme cases:
$$\begin{align}f(a)&=\left(a-\frac12\right)^2-\frac14>0 \Rightarrow a\in (-\infty,0)\cup (1,+\infty)\\
f(a)&=\left(a-\frac15\right)^2+\frac15>0 \Rightarrow a\in (-\infty,+\infty)\\
f(a)&=\left(a+\frac12\right)^2-\frac94>0 \Rightarrow a\in (-\infty,-2)\cup (1,+\infty)\end{align}$$
which can be observed from the Desmos graph too.
