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Beliebt Trigonometrie >

sec^2(θ)tan(θ)=2tan(θ),0<= θ<= 360

Lösung

sec^{2} (θ) tan (θ) = 2 tan (θ), 0^{\circ} \leq θ \leq 36 0^{\circ}

Lösung

θ = 0, θ = 18 0^{\circ}, θ = 36 0^{\circ}, θ = 13 5^{\circ}, θ = 22 5^{\circ}, θ = 4 5^{\circ}, θ = 31 5^{\circ}

Radianten

θ = 0, θ = π, θ = 2 π, θ = \frac{3 π}{4}, θ = \frac{5 π}{4}, θ = \frac{π}{4}, θ = \frac{7 π}{4}

Schritte zur Lösung

sec^{2} (θ) tan (θ) = 2 tan (θ), 0^{\circ} \leq θ \leq 36 0^{\circ}

Subtrahiere

2 tan (θ)

von beiden Seiten

sec^{2} (θ) tan (θ) - 2 tan (θ) = 0

Faktorisiere

sec^{2} (θ) tan (θ) - 2 tan (θ) : tan (θ) (sec (θ) + 2) (sec (θ) - 2)

sec^{2} (θ) tan (θ) - 2 tan (θ)

Klammere gleiche Terme aus

tan (θ)

= tan (θ) (sec^{2} (θ) - 2)

Faktorisiere

sec^{2} (θ) - 2 : (sec (θ) + 2) (sec (θ) - 2)

sec^{2} (θ) - 2

Wende Radikal Regel an:

a = (a)^{2}

2 = (2)^{2}

= sec^{2} (θ) - (2)^{2}

Wende Formel zur Differenz von zwei Quadraten an:

x^{2} - y^{2} = (x + y) (x - y)

sec^{2} (θ) - (2)^{2} = (sec (θ) + 2) (sec (θ) - 2)

= (sec (θ) + 2) (sec (θ) - 2)

= tan (θ) (sec (θ) + 2) (sec (θ) - 2)

tan (θ) (sec (θ) + 2) (sec (θ) - 2) = 0

Löse jeden Teil einzeln

tan (θ) = 0 or sec (θ) + 2 = 0 or sec (θ) - 2 = 0

tan (θ) = 0, 0 \leq θ \leq 36 0^{\circ} : θ = 0, θ = 18 0^{\circ}, θ = 36 0^{\circ}

tan (θ) = 0, 0 \leq θ \leq 36 0^{\circ}

Allgemeine Lösung für

tan (θ) = 0

tan (x)

Periodizitätstabelle mit

18 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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = 0 + 18 0^{\circ} n

θ = 0 + 18 0^{\circ} n

Löse

θ = 0 + 18 0^{\circ} n : θ = 18 0^{\circ} n

θ = 0 + 18 0^{\circ} n

0 + 18 0^{\circ} n = 18 0^{\circ} n

θ = 18 0^{\circ} n

θ = 18 0^{\circ} n

Lösungen für den Bereich

0 \leq θ \leq 36 0^{\circ}

θ = 0, θ = 18 0^{\circ}, θ = 36 0^{\circ}

sec (θ) + 2 = 0, 0 \leq θ \leq 36 0^{\circ} : θ = 13 5^{\circ}, θ = 22 5^{\circ}

sec (θ) + 2 = 0, 0 \leq θ \leq 36 0^{\circ}

Verschiebe

2

auf die rechte Seite

sec (θ) + 2 = 0

Subtrahiere

2

von beiden Seiten

sec (θ) + 2 - 2 = 0 - 2

Vereinfache

sec (θ) = - 2

sec (θ) = - 2

Allgemeine Lösung für

sec (θ) = - 2

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

θ = 13 5^{\circ} + 36 0^{\circ} n, θ = 22 5^{\circ} + 36 0^{\circ} n

θ = 13 5^{\circ} + 36 0^{\circ} n, θ = 22 5^{\circ} + 36 0^{\circ} n

Lösungen für den Bereich

0 \leq θ \leq 36 0^{\circ}

θ = 13 5^{\circ}, θ = 22 5^{\circ}

sec (θ) - 2 = 0, 0 \leq θ \leq 36 0^{\circ} : θ = 4 5^{\circ}, θ = 31 5^{\circ}

sec (θ) - 2 = 0, 0 \leq θ \leq 36 0^{\circ}

Verschiebe

2

auf die rechte Seite

sec (θ) - 2 = 0

Füge

2

zu beiden Seiten hinzu

sec (θ) - 2 + 2 = 0 + 2

Vereinfache

sec (θ) = 2

sec (θ) = 2

Allgemeine Lösung für

sec (θ) = 2

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

θ = 4 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n

θ = 4 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n

Lösungen für den Bereich

0 \leq θ \leq 36 0^{\circ}

θ = 4 5^{\circ}, θ = 31 5^{\circ}

Kombiniere alle Lösungen

θ = 0, θ = 18 0^{\circ}, θ = 36 0^{\circ}, θ = 13 5^{\circ}, θ = 22 5^{\circ}, θ = 4 5^{\circ}, θ = 31 5^{\circ}

Graph

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