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

Your approach is fine.
You already got
$$\bigg(2\cos x+(1-a)\bigg)^2-4a^2-(1-a)^2<0$$
which is equivalent to
$$\bigg(2\cos x+(1-a)\bigg)^2\lt 5a^2-2a+1\tag1$$
Since $5a^2-2a+1$ is always positive, $(1)$ is equivalent to
$$-\sqrt{5a^2-2a+1}\lt 2\cos x+1-a\lt \sqrt{5a^2-2a+1}$$
which is equivalent to
$$\frac 12\left(a-1-\sqrt{5a^2-2a+1}\right)\lt \cos x\lt\frac 12\left(a-1+\sqrt{5a^2-2a+1}\right)\tag2$$
The necessary and sufficient condition that $(2)$ holds for any $x\in\mathbb R$ is
$$\frac 12\left(a-1-\sqrt{5a^2-2a+1}\right)\lt -1\quad\text{and}\quad 1\lt \frac 12\left(a-1+\sqrt{5a^2-2a+1}\right),$$
i.e.
$$\sqrt{5a^2-2a+1}\gt a+1\quad \text{and}\quad  \sqrt{5a^2-2a+1}\gt -a+3\tag3$$ 
If $a\lt -1$, then 
$$(3)\iff 5a^2-2a+1\gt (-a+3)^2\iff a\in (-\infty, -2)\cup (1,\infty)$$
So, in this case, we have $a\in (-\infty,-2)$.
If $-1\le a\lt 3$, then
$$\begin{align}(3)&\iff 5a^2-2a+1\gt (a+1)^2\quad\text{and}\quad 5a^2-2a+1\gt (-a+3)^2\\\\&
\iff a\in (-\infty, -2)\cup (1,\infty)\end{align}$$
So, in this case, we have $a\in (1,3)$.
If $a\ge 3$, then
$$(3)\iff 5a^2-2a+1\gt (a+1)^2\iff a\in (-\infty,0)\cup (1,\infty)$$
So, in this case, we have $a\in [3,\infty)$.
Therefore, the range of $a$ is
$$a\in (-\infty,-2)\cup (1,3)\cup [3,\infty),$$
i.e.
$$a\in (-\infty, -2)\cup (1,\infty)$$

Another approach
First, let us prove the following lemma : 
Lemma 1 : 
If the inequality holds for any $x\in\mathbb R$, then $a$ has to, at least, satisfy $a\lt -2$ or $a\gt 1$.
Proof : 
The inequality 
$$\sin^2 x+a\cos x+a^2>1+\cos x$$
holds for $x=0$, so we have to have
$$a+a^2\gt 2,$$
i.e.
$$a\lt -2\quad\text{or}\quad a\gt 1\qquad\square$$
Next, let us prove the following lemma :
Lemma 2 : 
For any $a$ satisfying $a\lt -2$ or $a\gt 1$, the inequality holds for any $x\in\mathbb R$.
Proof : 
The inequality can be written as
$$t^2+(1-a)t-a^2\lt 0\tag4$$
where  $t=\cos x$.
So, we want to prove that
$$a\in (-\infty, -2)\cup (1,\infty)\implies \text{$(4)$ holds for any $t$ satisfying $|t|\le 1$}\tag5$$
Let $f(t)$ be the LHS of $(4)$, and let us consider the graph of $y=f(t)$.
If $a\in (-\infty, -2)\cup (1,\infty)$, then we have
$$f(-1)=-a^2+a\lt 0\qquad \text{and}\qquad f(1)=-a^2-a+2\lt 0$$
Since the graph of $y=f(t)$ is an upward parabola, it follows that if $a\in (-\infty, -2)\cup (1,\infty)$, then $f(t)\lt 0$ for any $t$ satisfying $|t|\le 1.\quad\square$
If follows from Lemma 1 and Lemma 2 that the range of $a$ is
$$a\in (-\infty, -2)\cup (1,\infty)$$
