cos(x)= 1/2 ,0<= x<= 2pi

Lösung

cos (x) = \frac{1}{2}, 0 \leq x \leq 2 π

x = \frac{π}{3}, x = \frac{5 π}{3}

Grad

x = 6 0^{\circ}, x = 30 0^{\circ}

Schritte zur Lösung

cos (x) = \frac{1}{2}, 0 \leq x \leq 2 π

Allgemeine Lösung für

cos (x) = \frac{1}{2}

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}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Lösungen für den Bereich

0 \leq x \leq 2 π

x = \frac{π}{3}, x = \frac{5 π}{3}

sec (θ) = 1

sin (θ) + cos (θ) = 0

- cos^{2} (θ) + sin^{2} (θ) = cos (θ) + 1

2 cos (x) = 0

cot (θ) = - 1