Problem 11.1. Find all values of the real parameter $a$ such that the equation

$$
a(\sin 2 x+1)+1=(a-3)(\sin x+\cos x)
$$

has a solution.

Emil Kolev
11.1. We write the given equation as

$$
2 a y^{2}-\sqrt{2}(a-3) y+1=0
$$

where

$$
y=\frac{\sqrt{2}}{2}(\sin x+\cos x)=\sin \left(x+45^{\circ}\right) \in[-1,1]
$$

It has a solution if and only if (1) has a solution in the interval $[-1,1]$.

For $a=0$ the equation (1) is linear and its root $y=-\frac{\sqrt{2}}{6}$ belongs to the interval $[-1,1]$.

Let $a \neq 0$ and set $f(y)=2 a y^{2}-\sqrt{2}(a-3) y+1$. We have $f(1)=0$ and $f(-1)=0$ for $a=-\frac{7 \sqrt{2}+8}{2}$ and $a=\frac{7 \sqrt{2}-8}{2}$, respectively. The quadratic polynomial $f(y)$ has exactly one root in $(-1,1)$ if and only if $f(-1) f(1)0 \\
& a f(1)>0 \\
& D=2 a^{2}-20 a+18 \geq 0 \Longleftrightarrow \\
& -10$ and $f\left(-\frac{1}{\sqrt{2}}\right)=2 a-2 \leq 0$, which implies that the equation $f(y)=0$ has a root in the interval $\left(-\frac{1}{\sqrt{2}}, 0\right) \subset(-1,1)$.

For $a \geq 9$ we have $f(0)=1>0$ and $f\left(\frac{1}{3 \sqrt{2}}\right)=\frac{2}{9}(9-a) \leq 0$, which shows that the equation $f(y)=0$ has a root in the interval $\left(0, \frac{1}{3 \sqrt{2}}\right) \subset(-1,1)$.

Hence the given equation has a solution if and only if $a \in(-\infty, 1] \cup[9, \infty)$.