Parametric equation with a trigonometric function in exponent I'm having trouble thinking of any solution or idea to solve this math problem. Any help will be appreciated.
$(a^2 - 1)*2^{-\sin^2 x} = a^2 - 4a + 3 $, 
$a = ?$ 
The equation should have real solutions.
 A: For $a=1$ the equation is satisfied for every $x$. Let's look for solutions different from $1$, so we can divide both sides by $a-1$, getting
$$
(a+1)2^{-{\sin^2x}}=a-3
$$
so
$$
2^{-{\sin^2x}}=\frac{a-3}{a+1} \tag{*}
$$
(note that $a=-1$ would lead to a contradiction).
Since $-1\le-{\sin^2x}\le0$, we have $1/2\le 2^{-{\sin^2x}} \le 1$.
Thus we need
$$
\frac{1}{2}\le \frac{a-3}{a+1}\le 1
$$
that is satisfied for $a\ge7$. Conversely, the equation (*) has solutions for every $a\ge7$. 
How to find $a\ge7$? We need
$$
\frac{1}{2}\le \frac{a+1-4}{a+1}\le 1
$$
that is
$$
\frac{1}{2}\le 1-\frac{4}{a+1}\le 1
$$
that is
$$
0\le\frac{4}{a+1}\le\frac{1}{2}
$$
Therefore $a+1>0$ and $8\le a+1$, that is, $a\ge7$.
A: $$a\in \lbrace [1,1] ,  [7, \infty) \rbrace \subset \mathbb {R}$$
if you graph it, with desmos (by setting $a=y$) for instance, indeed $a=1$ all $x$.  But with the equation so presented you can arrive at other ordered pairs $(x,a)$ such that $a \neq 1$.  Can someone account for this dilemma?
Allowing that $a \neq 1$ and solving it for $a$ in terms of $x$ produces
$$a(x)=3+\frac{4}{2^{{\sin^2 x}}-1} \ge 7 \quad \forall x$$
and for example 
$$a\left(\frac{n\pi}{2}\right)=7 \implies \left(\frac{n\pi}{2},7\right) \quad  \bigg {|} \quad n \in \mathbb {Z}$$
A: For $a=1$ $$(a^2 - 1).2^{-\sin^2 x} = a^2 - 4a + 3$$ is simply $$0=0$$ which is trivial so $x$ could be any real number. Otherwise we divide both sides by $(a-1)$ to get    $$(a+ 1).2^{-\sin^2 x} = a- 3$$ For $a=-1$ the equation turns into $$0=-4$$which has no solution. Otherwise we get  $2^{-\sin^2 x}=(a-3)/(a+1).$ Note that  $1/2\le 2^{-\sin^2 x}\le 1$. $ 1/2\le (a-3)/(a+1)\le 1$ has solutions on $a\in [7,\infty)$. Thus  we have solutions if $a=1$ or  $ a\in  [7,\infty) $.
