3. Find all values of the parameter $a$ for which the equation

$$
\frac{|\cos t-0.5|+|\sin t|-a}{\sqrt{3} \sin t-\cos t}=0
$$

has at least one solution on the interval $[0 ; \pi / 2]$. Specify the number of distinct solutions of this equation on the interval $[0 ; \pi / 2]$ for each found value of the parameter $a$. (12 points)
Solution. Let $x=\cos t, y=\sin t, x^{2}+y^{2}=1$, $x \in[0 ; 1], y \in[0 ; 1]$. Then the equation will have the form

$$
\frac{|x-0.5|+|y|-a}{\sqrt{3} y-x}=0
$$

$$
|x-0.5|+|y|-a=0, \sqrt{3} y-x \neq 0
$$

In the end, we have the system:

$\left\{\begin{array}{c}|x-0.5|+|y|=a, \\ \sqrt{3} y-x \neq 0, \\ x^{2}+y^{2}=1, x \in[0 ; 1], y \in[0 ; 1]\end{array}\right.$

Note that

$a>0$. The intersection point of the circle

$x^{2}+y^{2}=1, x \in[0 ; 1], y \in[0 ; 1]$, with the line $\sqrt{3} y- x=0$ has coordinates $(\sqrt{3} / 2 ; 1 / 2)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_b07d651c344f50283694g-17.jpg?height=860&width=897&top_left_y=438&top_left_x=1162)

Let's find the value of the parameter $a$ for which the square $|x-0.5|+|y|=a$ passes through the point $(1 ; 0)$. We have $a=0.5$. For $a>1.5$ the system has no solutions.

Answer: the equation has at least one solution for $a \in[0.5 ; 1.5]$, for $a \in[0.5 ; \sqrt{3} / 2] \cup(\sqrt{2}-0.5 ; 1.5]$ the equation has 1 solution, for $a \in(\sqrt{3} / 2 ; \sqrt{2}-0.5)$ the equation has 3 solutions, for $a=\sqrt{2}-0.5$ the equation has 2 solutions.
4b. Find the angle between the planes $BCD$ and $AFE$ (see the condition of problem 4a). (8 points)

## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_b07d651c344f50283694g-18.jpg?height=994&width=1650&top_left_y=517&top_left_x=226)

Introduce a coordinate system as shown in the figure. Then $A=(0 ; 0 ; 0), B=(0 ; 40 ; 0), E=$ $(40 ; 10 ; 40), F=(80 ; 40 ; 40), C=(40 ; 30 ; 0), D=(80 ; 0 ; 40)$. We have $\overrightarrow{BD}=\{80 ;-40 ; 40\}, \overrightarrow{BC}=$ $\{40 ;-10 ; 0\}, \overrightarrow{AE}=\{40 ; 10 ; 40\}, \overrightarrow{AF}=\{80 ; 40 ; 40\}$. Let $\vec{n} \perp \overrightarrow{BC}, \vec{n} \perp \overrightarrow{BD}, \vec{m} \perp \overrightarrow{AE}, \vec{m} \perp$ $\overrightarrow{AF}, \vec{n}=\{a ; b ; c\}, \vec{m}=\{d ; e ; f\}$. Then we have two systems of linear algebraic equations: $\left\{\begin{array}{c}40 a-10 b=0, \\ 80 a-40 b+40 c=0,\end{array} \quad\left\{\begin{array}{l}40 d+10 e+40 f=0, \\ 80 d+40 e+40 f=0 .\end{array}\right.\right.$ For the first system, we have $\left\{\begin{array}{l}b=4 a, \\ c=2 a,\end{array}\right.$ and as a normal vector to the plane $BCD$ we can choose the vector $\vec{n}=\{1 ; 4 ; 2\}$. For the second system, we have $\left\{\begin{array}{c}e=2 f, \\ d=-3 f / 2\end{array}\right.$, and as a normal vector to the plane $AFE$ we can choose the vector $\vec{m}=\{-3 ; 4 ; 2\}$. The angle between the planes is equal to the angle between the normals to these planes. Therefore, the cosine of the desired angle $\alpha$ can be found using the formula

$$
\cos \alpha=\frac{|\vec{n} \cdot \vec{m}|}{|\vec{n}| \cdot|\vec{m}|}=\frac{17}{\sqrt{21 \cdot 29}}=\frac{17 \sqrt{609}}{609}
$$

Answer: $\arccos \frac{17 \sqrt{609}}{609}$.