sin(θ)=(-sqrt(3))/2 ,0<= θ<= 4pi

Lösung

sin (θ) = \frac{- 3}{2}, 0 \leq θ \leq 4 π

θ = \frac{4 π}{3}, θ = \frac{5 π}{3}, θ = \frac{10 π}{3}, θ = \frac{11 π}{3}

Grad

θ = 24 0^{\circ}, θ = 30 0^{\circ}, θ = 60 0^{\circ}, θ = 66 0^{\circ}

Schritte zur Lösung

sin (θ) = \frac{- 3}{2}, 0 \leq θ \leq 4 π

Wende Bruchregel an:

\frac{- a}{b} = - \frac{a}{b}

sin (θ) = - \frac{3}{2}

Allgemeine Lösung für

sin (θ) = - \frac{3}{2}

sin (x)

Periodizitätstabelle mit

2 πn

Zyklus:

θ = \frac{4 π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

θ = \frac{4 π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Lösungen für den Bereich

0 \leq θ \leq 4 π

θ = \frac{4 π}{3}, θ = \frac{5 π}{3}, θ = \frac{10 π}{3}, θ = \frac{11 π}{3}

4 sin (θ) + 4 = \frac{- 3}{sin ( θ ) - 1}

cos (θ) = 0.707

sin (x) + sin^{2} (x) + sin^{3} (x) + sin^{4} (x) = 0

sin (9 x) - cos (x) = 0

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