Solve parameter function 
Solve $4^{\sin{x}}+m×2^{\sin{x}}+m^2-1=0$ over parameter $m$

I dunno what to process but I want to show some of my work at least. I try to solve for $m$ firstly so I get
\begin{equation}
 m_1=\frac{-2^{\sin{x}}+\sqrt{4-3×4^{\sin{x}}}}{2}\\m_2=\frac{-2^{\sin{x}}-\sqrt{4-3×4^{\sin{x}}}}{2}
\end{equation}
And then I try to find the range of $m_1$ and $m_2$ which give
\begin{equation}
\frac{-1+\sqrt{13}}{4}\geq m_1\geq -1\\ \frac{-1}{4}\geq m_2\geq \frac{-4-\sqrt{13}}{4}
\end{equation}
Because of $\Delta \geq 0$
Now I dunno what to do next. Please help
 A: $$4^{\sin x}+m 2^{\sin x}+m^2-1=0~~~~(1)$$
Here $x$ is the variable whose roots are to be obtained  and $m$ is a parameter. Let $2^{\sin x}=z \in [1/2,2]$, get
$$f(z)=z^2+mz+m^2-1=0~~~~(2)$$
We look for range of the parameter so that $z$ has at least one root in $[1/2,2]$
There are two cases: (2) exactly one root in $[1/2,2]$, (2) both roots in $[1/2,2]$.
Case (1): for this we require
$$f(1/2)f(2)\le 0 \implies (4m^2+2m-3)(m^2+2m+3) \le 0 \implies 4m^2+2m-3 \le 0$$ $$ \implies m \in [-\frac{\sqrt{13}+1}{4},\frac{\sqrt{13}-1}{4}]~~~~(3).$$
Note that the second quadratic in above is positive definite.
Case (2): For (1) to have both roots in $[1/2,2]$, we require $$ (a) B^2 \ge 4AC, (b) f(1/2) \ge 0, (c) f(2) \ge 0,  (d) 1/2 \le z_0=-\frac{B}{2A}=-m/2 \le 2$$
(a) gives $-2/\sqrt{3}\le  m\le 2/\sqrt{3}$
(b) gives $ m\in (-\infty,- \frac{\sqrt{13}+1}{4}) \cup (\frac{\sqrt{13}+1}{4}, \infty)$.
(c) gives $m\in (-\infty, \infty)$
(d) gives $-4 <m < -1$
The overlap of these cases gives $$-\frac{\sqrt{13}+1}{4} \le m  \le-1.~~~(4)$$
Finally, one (3) or two (4) physical roots of (*) can be obtained as $x=\sin^{-1}\ln z$, where
$$ \implies z=\frac{-m\pm \sqrt{4-3m^2}}{2},$$
using the restrictions on $m$ as given by (3) and (4), respectively in $x \in [0,2\pi].$
It may be noted that $(*)$ being a periodic Eqn. it will have INFINITELY many roots in the parametric regimes (3) and (4). But in $x \in [0,2\pi]$, (*) will have one or two roots.
A: $4^{\sin{x}}+m×2^{\sin{x}}+m^2-1=0$
Say, $y = 2^{\sin x}$ and so we have $y^2 + my + (m^2-1) = 0$ ...(i)
For equation to have any real solution, its discrimant has to be $\ge 0$ i.e.
$m^2 - 4(m^2-1) \ge 0 \implies - \frac{2}{\sqrt 3} \le m \le \frac{2}{\sqrt 3}$ ...(ii)
Now solving (i), $ (y + \frac{m}{2})^2 = 1 - \frac{3m^2}{4} \implies y = -\frac{m}{2} \pm \sqrt {1 - \frac{3m^2}{4}}$ ...(iii)
$y = 2^{\sin x} \gt 0\,$ has a minimum value of $\frac{1}{2}$ (when $\sin x = -1$) and a max value of $2$ (when $\sin x = 1$).
That leads to further constraint on (ii) and $ - \frac{2}{\sqrt 3} \le m \le \frac{\sqrt 13 - 1}{4}$.
The upper bound comes from the fact that $y \ge \frac{1}{2}$. Further, any greater value of $m$ will mean modulus of  $-\frac{m}{2}$ will increase (which is negative) whereas the value inside the square root will decrease. So $y$ will decrease which does not give a solution.
EDIT: Just some further remarks on the lower bound.
The lower bound $-\frac{2}{\sqrt 3}$ gives value of $y$ as $\frac{1}{\sqrt 3} (\gt \frac {1}{2})$ which works; $m = -\frac{\sqrt{13}+1}{4} (\gt - \frac{2}{\sqrt 3})$ gives $y = \frac {1}{2}$ but is not the lowest bound of $m$.
