Bulgarian Mathematical Competitions

Solution:
Set $t=\sin x$ and $g(t)=\frac{t^{2}-a}{t^{3}-\left(a^{2}+2\right) t+2}$.
If the numerator and the denominator of $g$ have a common root then $a \geq 0$ and $t= \pm \sqrt{a}$. If $t=-\sqrt{a}$ we obtain $\sqrt{a}\left(a^{2}-a+2\right)=-2$, which is impossible since $a^{2}-a+2>0$ for every $a$. If $t=\sqrt{a}$ we obtain $\sqrt{a}(a(a-1)+2)=2$ and it is easy to see that $a=1$ is the only solution of this equation (for $a \in[0,1)$ the left hand side is less than 2, and for $a>1$ it is greater than 2). For $a=1$ we have
$$g(t)=\frac{t^{2}-1}{t^{3}-3 t+2}=\frac{t+1}{(t-1)(t+2)} \leq 0$$
for every $t \in[-1,1)$ which implies that this value of $a$ is not a solution.

We now look for $a \neq 1$ such that the equation $g(t)=c$, i.e.
$$h(t)=c\left(t^{3}-\left(a^{2}+2\right) t+2\right)+a-t^{2}=0$$
has a solution in the interval $[-1,1]$ for every $c \in\left[\frac{1}{2}, 2\right]$. For $c \geq \frac{1}{2}$ we have
$$h(-1)=c a^{2}+a+3 c-1 \geq \frac{a^{2}}{2}+a+\frac{1}{2}=\frac{(a+1)^{2}}{2}$$
whence $h(-1) \geq 0$ for every $a$. Also, $h'(t)=3 c t^{2}-2 t-c\left(a^{2}+2\right)$ and therefore $h'(1)=c-c a^{2}-2 \leq 0$ for $c \in\left[\frac{1}{2}, 2\right]$ and every $a$. Hence the equation $h'(t)=0$ has real roots $t_{1}$ and $t_{2}$ such that $t_{1} \leq 1<t_{2}$.
Thus the function $h(t)$ is decreasing in the interval $\left[t_{1}, t_{2}\right] \ni 1$ and increasing in the interval $\left(-\infty, t_{1}\right]$. Since $h(-1) \geq 0$, it is easy to see that $h(t)=0$ has a solution in the interval $[-1,1]$ for every $c \in\left[\frac{1}{2}, 2\right]$ if and only if
$$h(1)=(a-1)(1-c(1+a)) \leq 0$$
for every such $c$. For $a>1$ this inequality is satisfied since $1-\frac{1}{2}(1+a)<0$, and for $a<1$ it is equivalent to $1-2(1+a) \geq 0$, i.e. $a \leq-\frac{1}{2}$.
Finally, the required values of $a$ are $a \in\left(-\infty,-\frac{1}{2}\right] \cup(1,+\infty)$.

Second solution:
Using the same reasoning as above we reduce the problem to finding of those $a$ such that $\left[\frac{1}{2}, 2\right] \subset g([-1,1])$.
Set $h(t)=t^{3}-\left(a^{2}+2\right) t+2$. Then $h'(t)=3 t^{2}-\left(a^{2}+2\right)$ and for $a^{2} \geq 1$ it follows that $h'(t)<0$ for $t \in(-1,1)$. Therefore $h(t)$ is a strictly decreasing continuous function in the interval $(-1,1]$. Since
$$h(1)=1-a^{2} \leq 0<a^{2}+3=h(-1)$$
it follows that $h(t)$ has a unique zero $t_{0} \in(-1,1]$.

We now consider several cases.

Case 1. Let $a>1$. Then $t_{0} \neq 1, \lim _{t \rightarrow t_{0}+0} g(t)=+\infty$ and $g(1)=\frac{1-a}{1-a^{2}}=\frac{1}{1+a}<\frac{1}{2}$. Since the function $g(t)$ is decreasing and continuous in $\left(t_{0}, 1\right]$, it follows that
$$\left[\frac{1}{2}, 2\right] \subset\left[\frac{1}{2},+\infty\right) \subset g\left(\left(t_{0}, 1\right]\right) \subset g([-1,1])$$

Case 2. Let $a \leq-1$. Now $\lim _{t \rightarrow t_{0}-0} g(t)=+\infty$ and $g(-1)=\frac{1-a}{a^{2}+3} \leq \frac{1}{2}$ (the last inequality is equivalent to $(a+1)^{2} \geq 0$ ). Then
$$\left[\frac{1}{2}, 2\right] \subset\left[\frac{1}{2},+\infty\right) \subset g\left(\left[-1, t_{0}\right)\right) \subset g([-1,1])$$

Case 3. Let $a=1$. Then we see as in the first solution that $g(t) \leq 0$ for $t \in[-1,1)$.

Case 4. Let $a \in(-1,1)$. We first check that $h(t)>0$ for $t \in[-1,1]$, which implies that $g(t)$ is a continuous function in that interval.
Indeed, if $t \in[-1,0]$, then $h(t) \geq t^{3}+2>0$ and if $t \in(0,1]$, then $h(t)>t^{3}-3 t+2=(t-1)^{2}(t+2) \geq 0$.

Case 4.1. If $a \in\left(-1,-\frac{1}{2}\right]$, then $g(-1)=\frac{1-a}{a^{2}+3}<\frac{1}{2}, g(1)=\frac{1}{1+a} \geq 2$ and therefore $\left[\frac{1}{2}, 2\right] \subset g([-1,1])$.

Case 4.2. If $a \in\left(-\frac{1}{2}, 1\right)$, we shall show that $g(t)<2$ for every $t \in[-1,1]$. Since $h(t)>0$ in this interval, we have to check that
$$t^{2}<2\left(t^{3}-\left(a^{2}+2\right) t+2\right)$$
i.e.
$$m(t)=2 t^{3}-t^{2}-2\left(a^{2}+2\right) t+a+4>0,\ t \in[-1,1]$$
We have
$$m'(t)=6 t^{2}-2 t-2\left(a^{2}+2\right) \leq 6 t^{2}-2 t-4=(6 t+4)(t-1)$$
and therefore $m'(t) \leq 0$ in the interval $\left[-\frac{2}{3}, 1\right]$. Hence $m(t)$ is a decreasing function in this interval which implies that
$$m(t) \geq m(1)=-2 a^{2}+a+1=(1+2 a)(1-a)>0$$
for any $t \in\left[-\frac{2}{3}, 1\right]$. On the other hand, we have
$$m(t) \geq 2 t^{3}-t^{2}+a+4 \geq-2-1-\frac{1}{2}+4>0$$
for $t \in[-1,0]$, which completes the proof of (1).

The cases considered above cover all possibilities for $a$ and we conclude that the answer is $a \in\left(-\infty,-\frac{1}{2}\right] \cup(1,+\infty)$.