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Beliebt Trigonometrie >

cos^2(θ)=0.1831

Lösung

cos^{2} (θ) = 0.1831

Lösung

θ = 1.12862 \dots + 2 πn, θ = 2 π - 1.12862 \dots + 2 πn, θ = 2.01296 \dots + 2 πn, θ = - 2.01296 \dots + 2 πn

+1

Grad

θ = 64.66551 \dots^{\circ} + 36 0^{\circ} n, θ = 295.33448 \dots^{\circ} + 36 0^{\circ} n, θ = 115.33448 \dots^{\circ} + 36 0^{\circ} n, θ = - 115.33448 \dots^{\circ} + 36 0^{\circ} n

Schritte zur Lösung

cos^{2} (θ) = 0.1831

Löse mit Substitution

cos^{2} (θ) = 0.1831

Angenommen:

cos (θ) = u

u^{2} = 0.1831

u^{2} = 0.1831 : u = 0.1831, u = - 0.1831

u^{2} = 0.1831

Für

x^{2} = f (a)

sind die Lösungen

x = f (a), - f (a)

u = 0.1831, u = - 0.1831

Setze in

u = cos (θ)

ein

cos (θ) = 0.1831, cos (θ) = - 0.1831

cos (θ) = 0.1831, cos (θ) = - 0.1831

cos (θ) = 0.1831 : θ = arccos (0.1831) + 2 πn, θ = 2 π - arccos (0.1831) + 2 πn

cos (θ) = 0.1831

Wende die Eigenschaften der Trigonometrie an

cos (θ) = 0.1831

Allgemeine Lösung für

cos (θ) = 0.1831

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

θ = arccos (0.1831) + 2 πn, θ = 2 π - arccos (0.1831) + 2 πn

θ = arccos (0.1831) + 2 πn, θ = 2 π - arccos (0.1831) + 2 πn

cos (θ) = - 0.1831 : θ = arccos (- 0.1831) + 2 πn, θ = - arccos (- 0.1831) + 2 πn

cos (θ) = - 0.1831

Wende die Eigenschaften der Trigonometrie an

cos (θ) = - 0.1831

Allgemeine Lösung für

cos (θ) = - 0.1831

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

θ = arccos (- 0.1831) + 2 πn, θ = - arccos (- 0.1831) + 2 πn

θ = arccos (- 0.1831) + 2 πn, θ = - arccos (- 0.1831) + 2 πn

Kombiniere alle Lösungen

θ = arccos (0.1831) + 2 πn, θ = 2 π - arccos (0.1831) + 2 πn, θ = arccos (- 0.1831) + 2 πn, θ = - arccos (- 0.1831) + 2 πn

Zeige Lösungen in Dezimalform

θ = 1.12862 \dots + 2 πn, θ = 2 π - 1.12862 \dots + 2 πn, θ = 2.01296 \dots + 2 πn, θ = - 2.01296 \dots + 2 πn

Graph

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