sin(θ)= 1/2 ,0<= θ<= 4pi

Lösung

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

θ = \frac{π}{6}, θ = \frac{5 π}{6}, θ = \frac{13 π}{6}, θ = \frac{17 π}{6}

Grad

θ = 3 0^{\circ}, θ = 15 0^{\circ}, θ = 39 0^{\circ}, θ = 51 0^{\circ}

Schritte zur Lösung

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

Allgemeine Lösung für

sin (θ) = \frac{1}{2}

sin (x)

Periodizitätstabelle mit

2 πn

Zyklus:

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Lösungen für den Bereich

0 \leq θ \leq 4 π

θ = \frac{π}{6}, θ = \frac{5 π}{6}, θ = \frac{13 π}{6}, θ = \frac{17 π}{6}

3 - cosh (x) = 0

1 (sin (x) - 0.3 π) + 2 = 0

8 - 32 cos^{2} (t) = 0

tan (x) = \frac{12}{9}

2 cos^{2} (x) - cos (x) + 1 = 0