## Task B-3.7.

For which values of the real number $t$ does the equation

$$
x^{2}+\frac{1}{\sqrt{\cos t}} 2 x+\frac{1}{\sin t}=2 \sqrt{2}
$$

have exactly one solution?
## Solution.

Obviously, it must be $\cos t>0$ and $\sin t \neq 0$.

For the quadratic equation to have one solution, its discriminant must be equal to 0, that is,

$$
\left(\frac{2}{\sqrt{\cos t}}\right)^{2}-4\left(\frac{1}{\sin t}-2 \sqrt{2}\right)=0
$$

By simplifying, we get $\frac{4}{\cos t}-\frac{4}{\sin t}=-8 \sqrt{2}$.

Multiplying by the denominator and rearranging, we have $\cos t-\sin t=\sqrt{2} \sin 2 t$.

Multiplying by $\frac{\sqrt{2}}{2}$ and applying the addition formula: $\sin \left(\frac{\pi}{4}-t\right)=\sin 2 t$.

Solving this equation, we get

$$
\begin{array}{rll}
\frac{\pi}{4}-t=2 t+2 k \pi & \text { or } & \frac{\pi}{4}-t=\pi-2 t+2 k \pi \\
3 t=\frac{\pi}{4}-2 k \pi & \text { or } & t=\pi-\frac{\pi}{4}+2 k \pi \\
t=\frac{\pi}{12}-\frac{2 k \pi}{3} & \text { or } & t=\frac{3 \pi}{4}+2 k \pi
\end{array}
$$

For none of the obtained solutions is $\sin t$ zero, but only the solutions $t=\frac{\pi}{12}+2 k \pi, k \in \mathbb{Z}$ have a positive cosine, so these are the desired values.

## SCHOOL/CITY COMPETITION IN MATHEMATICS

4th grade - high school - B variant
January 5, 2020.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH IN AN APPROPRIATE MANNER.