1. Find all values of $m$ in the interval ($-1,1$) for which the equation

$$
4^{\sin x}+m 2^{\sin x}+m^{2}-1=0
$$

has solutions.
Solution. Let's set $y=2^{\sin x}$. Thus we obtain the equation

$$
y^{2}+m y+m^{2}-1=0
$$

The discriminant $D=4-3 m^{2}$ of equation (1) for $m \in(-1,1)$ is positive, which means that for every $m \in(-1,1)$ equation (1) has real solutions. The solutions are

$$
y_{1 / 2}=\frac{1}{2}\left(-m \pm \sqrt{4-3 m^{2}}\right)
$$

From $y=2^{\sin x}>0$ it follows that the solution can only be $y_{1}=\frac{1}{2}\left(-m+\sqrt{4-3 m^{2}}\right)$. For $m \in(-1,1)$ we have $4-3 m^{2}>m^{2}$, which means that $y_{1}>0$. Now we have:

$$
2^{\sin x}=\frac{1}{2}\left(-m+\sqrt{4-3 m^{2}}\right), \text { i.e. } \sin x=\log _{2}\left[\frac{1}{2}\left(-m+\sqrt{4-3 m^{2}}\right)\right]
$$

or

$$
\sin x=\log _{2}\left(-m+\sqrt{4-3 m^{2}}\right)-1
$$

For equation (2) to have a solution, it is necessary that

$$
-1 \leq \log _{2}\left(-m+\sqrt{4-3 m^{2}}\right)-1 \leq 1
$$

i.e.

$$
1 \leq -m+\sqrt{4-3 m^{2}} \leq 4
$$

which is satisfied for every $m \in\left[\frac{-1-\sqrt{13}}{4}, \frac{-1+\sqrt{13}}{4}\right]$.

Therefore, the given equation has a solution for every $m \in\left(-1, \frac{-1+\sqrt{13}}{4}\right]$.