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Beliebt >

8cos^2(θ)=2

Lösung

8 cos^{2} (θ) = 2

Lösung

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

+1

Grad

θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n, θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n

Schritte zur Lösung

8 cos^{2} (θ) = 2

Löse mit Substitution

8 cos^{2} (θ) = 2

Angenommen:

cos (θ) = u

8 u^{2} = 2

8 u^{2} = 2 : u = \frac{1}{2}, u = - \frac{1}{2}

8 u^{2} = 2

Teile beide Seiten durch

8

8 u^{2} = 2

Teile beide Seiten durch

8

\frac{8 u ^{2}}{8} = \frac{2}{8}

Vereinfache

u^{2} = \frac{1}{4}

u^{2} = \frac{1}{4}

Für

x^{2} = f (a)

sind die Lösungen

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

u = \frac{1}{4}, u = - \frac{1}{4}

\frac{1}{4} = \frac{1}{2}

\frac{1}{4}

Wende Radikal Regel an:

angenommen

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Faktorisiere die Zahl:

4 = 2^{2}

= 2^{2}

Wende Radikal Regel an:

2^{2} = 2

= 2

= \frac{1}{2}

Wende Regel an

1 = 1

= \frac{1}{2}

- \frac{1}{4} = - \frac{1}{2}

- \frac{1}{4}

Vereinfache

\frac{1}{4} : \frac{1}{2}

\frac{1}{4}

Wende Radikal Regel an:

angenommen

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Faktorisiere die Zahl:

4 = 2^{2}

= 2^{2}

Wende Radikal Regel an:

2^{2} = 2

= 2

= \frac{1}{2}

= - \frac{1}{2}

Wende Regel an

1 = 1

= - \frac{1}{2}

u = \frac{1}{2}, u = - \frac{1}{2}

Setze in

u = cos (θ)

ein

cos (θ) = \frac{1}{2}, cos (θ) = - \frac{1}{2}

cos (θ) = \frac{1}{2}, cos (θ) = - \frac{1}{2}

cos (θ) = \frac{1}{2} : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

cos (θ) = \frac{1}{2}

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

cos (θ) = - \frac{1}{2} : θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

cos (θ) = - \frac{1}{2}

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

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

Kombiniere alle Lösungen

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

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