[tex]\displaystyle\\
1)\\\\
S = 2^{50}-C_{50}^1 \cdot 2^{49}\cdot 3+C_{50}^2 \cdot 2^{48}\cdot 3^2- \cdots -C_{50}^{49} \cdot 2 \cdot 3^{49} +3^{50}\\\\
\text{Aceasta este o dezvoltare a binomului lui Newton}\\\\
S = (2-3)^{50} = (-1)^{50}=(1)^{50}=\boxed{\bf 1}\\\\
\text{Semnul minus a disparut deoarece exponentul este un numar par.}[/tex]
[tex]\displaystyle\\
2)\\\\
\sin3x = - \frac{1}{2}~~~~~x\in [0,~ 1\pi ] \\\\
\text{Stim ca: }~~~\boxed{\sin30^o = \frac{1}{2}}\\\\
\text{Sinusul este negativ in cadranele 3 si 4}\\\\
\text{Rezulta ca avem 2 solutii.}\\\\
\text{Solutia 1 (in cadranul 3:)}\\\\
\sin3x = \sin(180^o + 30^o) = -\sin30^o = -\frac{1}{2}\\\\
\Longrightarrow~~~3x = 180^o+30^o = 210^o\\\\
\Longrightarrow~~~x_1 = \frac{210^o}{3} = \boxed{\bf70^o}
[/tex]
[tex]\displaystyle\\
\text{Solutia 2 (in cadranul 4:)}\\\\
\sin3x = \sin(360^o - 30^o) = -\sin30^o = -\frac{1}{2}\\\\
\Longrightarrow~~~3x = 360^o - 30^o = 330^o\\\\
\Longrightarrow~~~x_2 = \frac{330^o}{3} = \boxed{\bf 110^o} \\\\
[/tex]
