siehe PDF-Datei Seminar2
a) Alle reellen Lösungen!
cos 3 x = 2 5 {\displaystyle \!\cos {3x}={\frac {2}{5}}\,}
t a n 2 x = c o s x {\displaystyle \!tan2x=cosx\,}
sin ( π 6 ) ⋅ cos 2 ( x ) + 1 6 ⋅ sin 2 ( x ) = 1 2 − 1 3 ⋅ sin 2 ( x ) {\displaystyle \sin \left({\frac {\pi }{6}}\right)\cdot \cos ^{2}\left(x\right)+{\frac {1}{6}}\cdot \sin ^{2}\left(x\right)={\frac {1}{2}}-{\frac {1}{3}}\cdot \sin ^{2}\left(x\right)}
sin ( x ) ⋅ 1 2 + cos ( x ) ⋅ 1 2 = 3 4 {\displaystyle \sin(x)\cdot {\sqrt {\frac {1}{2}}}+\cos \left(x\right)\cdot {\sqrt {\frac {1}{2}}}={\sqrt {\frac {3}{4}}}}
b) Lösungen in [ 5 π 4 ; 9 π 4 ] {\displaystyle [{\frac {5\pi }{4}};{\frac {9\pi }{4}}]}
c o s ( 2 x − π 2 ) = − 1 2 {\displaystyle cos\left(2x-{\frac {\pi }{2}}\right)=-{\frac {1}{2}}}
c) Lösungen in [ 0 ; 2 π [ {\displaystyle [0;2\pi [}
c o s 2 x = 1 c o s x − 1 {\displaystyle cos2x={\frac {1}{cosx}}-1}
Bestimmen Sie alle Lösungen!
a) 1 x 1 + 3 x 2 + 2 x 3 = 24 {\displaystyle \!1x_{1}+3x_{2}+2x_{3}=24} 2 x 1 + 1 x 2 + 5 x 3 = 31 {\displaystyle \!2x_{1}+1x_{2}+5x_{3}=31} 3 x 1 + 4 x 2 + 2 x 3 = 40 {\displaystyle \!3x_{1}+4x_{2}+2x_{3}=40}
b) 1 x 1 + 2 x 2 + 1 x 3 = 1 {\displaystyle \!1x_{1}+2x_{2}+1x_{3}=1} 2 x 1 + 3 x 2 + 1 x 3 = 1 {\displaystyle \!2x_{1}+3x_{2}+1x_{3}=1} 1 x 1 + 3 x 2 + 2 x 3 = 0 {\displaystyle \!1x_{1}+3x_{2}+2x_{3}=0}
c) 1 x 1 + 2 x 2 + 1 x 3 = 1 {\displaystyle \!1x_{1}+2x_{2}+1x_{3}=1} 2 x 1 + 3 x 2 + 1 x 3 = 1 {\displaystyle \!2x_{1}+3x_{2}+1x_{3}=1} 1 x 1 + 3 x 2 + 2 x 3 = 2 {\displaystyle \!1x_{1}+3x_{2}+2x_{3}=2}
d) 1 x 1 + 2 x 2 + 1 x 3 = 1 {\displaystyle \!1x_{1}+2x_{2}+1x_{3}=1} 2 x 1 + 4 x 2 + 2 x 3 = 2 {\displaystyle \!2x_{1}+4x_{2}+2x_{3}=2} − x 1 − 2 x 2 − 1 x 3 = − 1 {\displaystyle \!-x_{1}-2x_{2}-1x_{3}=-1}