cos(θ)=1,0<= θ<= 180

Lösung

cos (θ) = 1, 0^{\circ} \leq θ \leq 18 0^{\circ}

θ = 0

Schritte zur Lösung

cos (θ) = 1, 0^{\circ} \leq θ \leq 18 0^{\circ}

Allgemeine Lösung für

cos (θ) = 1

cos (x)

Periodizitätstabelle mit

36 0^{\circ} n

Zyklus:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = 0 + 36 0^{\circ} n

θ = 0 + 36 0^{\circ} n

Löse

θ = 0 + 36 0^{\circ} n : θ = 36 0^{\circ} n

θ = 0 + 36 0^{\circ} n

0 + 36 0^{\circ} n = 36 0^{\circ} n

θ = 36 0^{\circ} n

θ = 36 0^{\circ} n

Lösungen für den Bereich

0 \leq θ \leq 18 0^{\circ}

θ = 0

sin (x + 3 7^{\circ}) = cos (2 x + 8)

cos^{2} (5 x) = 1

3 sin^{2} (x) = 9 cos^{2} (x)

3 cos^{2} (x) - 3 = sin (x)

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