<s>[INST] 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
 [/INST] Given task allows suitable algebraic approach, because it is equivalent to the system of
\begin{cases}
\cos x =t\\
t\in[-1,1]\\
t^2+(1-a)t-a^2<0,\tag1 
\end{cases}
wherein the parabola $P(t,a) = t^2+(1-a)t-a^2$ has the positive discriminant 
$$D=(1-a)^2+4a^2.\tag2$$
The roots of $P(t,a)$ are
$$t_1 = \dfrac{a-1 - \sqrt D}2,\quad t_2 = \dfrac{a-1 + \sqrt D}2\tag3$$
wherein
$$P(t,a) = (t-t_1)(t-t_2)\tag4$$
is negative iff
$$t\in(t_1,t_2).\tag5$$
Condition $(5)$ is satisfied for all $t\in[-1,1]$ iff
$\mathbf{[-1,1] \subset (t_1,t_2)},$
or $t_1<-1$ and $t_2>1.$
This leads to the system in the form of
\begin{cases}
a-1-\sqrt{(a-1)^2+4a^2} < -2\\[4pt]
a-1 +\sqrt{(a-1)^2+4a^2} > 2,\tag6
\end{cases}
$$\sqrt{(a-1)^2+4a^2} > \max(a+1,\,3-a).\tag7$$
Since
$$\max(a+1,3-a) =
\begin{cases}
3-a,\quad\text{if}\quad a\in(-\infty,1)\\[4pt]
a+1,\quad\text{if}\quad a\in[1,\infty)\tag8
\end{cases}$$
is positive, then the inequality $(7)$ can be presented in the form of
$$\genfrac{[}{.}{0}{0}
{a\in(-\infty,1),\quad \sqrt{5a^2-2a+1} > 3-a}
{a\in[1,+\infty),\quad \sqrt{5a^2-2a+1} > a+1.}\tag9$$
Then
$$\genfrac{[}{.}{0}{0}
{a\in(-\infty,1),\quad 5a^2-2a+1 > a^2-6a+9}
{a\in[1,+\infty),\quad 5a^2-2a+1 > a^2+2a+1,}$$
$$\genfrac{[}{.}{0}{0}
{a\in(-\infty,1),\quad a^2+a-2 > 0}
{a\in[1,+\infty),\qquad\,\, a^2-a > 0,}$$
$$\genfrac{[}{.}{0}{0}
{a\in(-\infty,1),\quad (a+2)(a-1) > 0}
{a\in[1,+\infty),\qquad\,\, a(a-1) > 0\quad,}\tag{10}$$
$$\color{green}{\mathbf{a\in (-\infty,-2) \cup (1,+\infty).}}\tag{11}$$
Solution of $(7)$ can be illustrated, using Wolfram Alpha.

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