2sin(x)=((4sin(x)-cos(x)))/2

Lösung

2 sin (x) = \frac{( 4 sin ( x ) - cos ( x ) )}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Grad

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Schritte zur Lösung

2 sin (x) = \frac{( 4 sin ( x ) - cos ( x ) )}{2}

Subtrahiere

\frac{4 sin ( x ) - cos ( x )}{2}

von beiden Seiten

\frac{cos ( x )}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

cos (x) = 0

Allgemeine Lösung für

cos (x) = 0

cos (x)

Periodizitätstabelle mit

2 πn

Zyklus:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x + 60) \cdot cos (x - 60) = \frac{1}{2}

\frac{cos ( 2 x + 9 )}{3} = \frac{1}{2}

sin^{6} (x) - cos^{6} (x) = 0

- sin (2 x) sin (x) + cos (2 x) cos (x) = 0

sin (2 x) + cos (2 x) - 1 = 0