<s>[INST] 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
 [/INST] $4^{\sin{x}}+m×2^{\sin{x}}+m^2-1=0$
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 $m$.
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