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.

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, $a\ge7$.
