[tex]Ne\ putem\ folosi\ de\ formul\ cosinusului:\\
Daca\ avem\ doi\ vectori\ \vec{a}=u_1\cdot\vec{i}+v_1\cdot\vec{j}\ si\ \vec{b}=u_2\cdot \vec{i}+v_2\cdot {j}\ atunci\\
cosinusul\ dintre\ cei\ doi\ vectori\ este\ egal\ cu:\\
\\
\boxed{cos\ \textless \ (\vec{a}, \vec{b})=\frac{u_1\cdot u_2+v_1\cdot v_2}{\sqrt{u_1^2+v_1^2}\cdot \sqrt{u_2^2+v_2^2}}}\\
Folosim\ aceasta\ formula:\\
\\
\cos \frac{\pi}{4}=\frac{2m\sqrt3+2m\sqrt3}{\sqrt{12+4m^2}\cdot
\sqrt{m^2+3}}\\
\\
[/tex]
[tex]\frac{\sqrt2}{2}=\frac{4m\sqrt3}{\sqrt{4(m^2+3)}\cdot \sqrt{m^2+3}}\\
\\
\frac{\sqrt2}{2}=\frac{4m\sqrt3}{2\sqrt{m^2+3}\cdot \sqrt{m^2+3}}\\
\\
\frac{\sqrt2}{2}=\frac{4m\sqrt3}{2m^2+6}\\
2\sqrt2m^2+6\sqrt2=8m\sqrt3\\
2\sqrt2m^2-8m\sqrt3+6\sqrt2=0\\
\Delta=192-4\cdot2\sqrt2\cdot6\sqrt2\\
\Delta=192-96\\
\Delta=96\Rightarrow \sqrt{\Delta}}=4\sqrt6\\
\\
m_1=\frac{8\sqrt3+4\sqrt6}{4\sqrt2}=\frac{4(2\sqrt3+\sqrt6)}{4\sqrt2}=\frac{2\sqrt3+\sqrt6}{\sqrt2}=\frac{2\sqrt6+2\sqrt3}{2}=\sqrt6+\sqrt3\\
[/tex]
[tex]Analog:m_2=\sqrt6-\sqrt3\\
S:m\in\{\sqrt6-\sqrt3,\sqrt6+\sqrt3\}[/tex]
