(2sqrt(3))/3 =sec(θ)

Lösung

\frac{2 3}{3} = sec (θ)

θ = \frac{π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Grad

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n

Schritte zur Lösung

\frac{2 3}{3} = sec (θ)

Tausche die Seiten

sec (θ) = \frac{2 3}{3}

Allgemeine Lösung für

sec (θ) = \frac{2 3}{3}

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}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

4 (sin (x))^{2} - 15 sin (x) - 4 = 0

1 + sin (θ) = 1 + sin (- θ)

cos (2 x) = 2 + 3 sin (x), \frac{π}{2} \leq x \leq \frac{3 π}{2}

tan (θ) = (\frac{3}{8})

29 = \frac{1}{2} (5.6) (7.1) sin (x)