cos(θ)= 1/2 ,0<= θ<= 4pi

Lösung

cos (θ) = \frac{1}{2}, 0 \leq θ \leq 4 π

θ = \frac{π}{3}, θ = \frac{5 π}{3}, θ = \frac{7 π}{3}, θ = \frac{11 π}{3}

Grad

θ = 6 0^{\circ}, θ = 30 0^{\circ}, θ = 42 0^{\circ}, θ = 66 0^{\circ}

Schritte zur Lösung

cos (θ) = \frac{1}{2}, 0 \leq θ \leq 4 π

Allgemeine Lösung für

cos (θ) = \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}

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

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

Lösungen für den Bereich

0 \leq θ \leq 4 π

θ = \frac{π}{3}, θ = \frac{5 π}{3}, θ = \frac{7 π}{3}, θ = \frac{11 π}{3}

sinh^{2} (x) = 2 sinh (x) cosh (x)

tan (3 x) cot (x + 4 0^{\circ}) = 1

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

4 sin (θ) = 23

sin (x) = 2 sin (x)