cos(θ)=-1/(sqrt(2))

Lösung

cos (θ) = - \frac{1}{2}

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

Schritte zur Lösung

cos (θ) = - \frac{1}{2}

Vereinfache

- \frac{1}{2} : - \frac{2}{2}

- \frac{1}{2}

Multipliziere mit dem Konjugat

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

Wende Radikal Regel an:

a a = a

22 = 2

= 2

= - \frac{2}{2}

cos (θ) = - \frac{2}{2}

Allgemeine Lösung für

cos (θ) = - \frac{2}{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{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

3 sin (2 t) + 4 = 1

so l v e f or x, cos (x) = 0

cos (x + \frac{π}{6}) - cos (x - \frac{π}{6}) = 1

tan (2 θ) + 2 sin (θ) = 0

2 cos^{2} (x) + 7 sin (x) = 5