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Beliebt Trigonometrie >

sec^2(θ)=2

Lösung

sec^{2} (θ) = 2

Lösung

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

+1

Grad

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

Schritte zur Lösung

sec^{2} (θ) = 2

Löse mit Substitution

sec^{2} (θ) = 2

Angenommen:

sec (θ) = u

u^{2} = 2

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

u^{2} = 2

Für

x^{2} = f (a)

sind die Lösungen

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

u = 2, u = - 2

Setze in

u = sec (θ)

ein

sec (θ) = 2, sec (θ) = - 2

sec (θ) = 2, sec (θ) = - 2

sec (θ) = 2 : θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

sec (θ) = 2

Allgemeine Lösung für

sec (θ) = 2

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

θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

sec (θ) = - 2 : θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

sec (θ) = - 2

Allgemeine Lösung für

sec (θ) = - 2

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

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

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

Kombiniere alle Lösungen

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

Graph

Beliebte Beispiele

sin^2(θ)=8-2cos^2(θ)

sin^{2} (θ) = 8 - 2 cos^{2} (θ)

10sin(x)+16=3sin(x)+9

10 sin (x) + 16 = 3 sin (x) + 9

2 sin (5 θ) = 0

solvefor x,sin^2(x)=1

so l v e f or x, sin^{2} (x) = 1

tan^4(x)-2=tan^2(x)+sec^2(x)

tan^{4} (x) - 2 = tan^{2} (x) + sec^{2} (x)