Question

Putin ajutooor!!! Ambele va rog!!

Answer 1

   
[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]