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Beliebt Trigonometrie >

cos(θ)>= (sqrt(3))/2 ,2pi>θ>= 0

Lösung

cos (θ) \geq \frac{3}{2}, 2 π > θ \geq 0

Lösung

0 \leq θ \leq \frac{π}{6} or 2 π - \frac{π}{6} \leq θ < 2 π

+2

Intervall-Notation

[0, \frac{π}{6}] \cup [2 π - \frac{π}{6}, 2 π)

Dezimale

0 \leq θ \leq 0.52359 \dots or 5.75958 \dots \leq θ < 6.28318 \dots

Schritte zur Lösung

cos (θ) \geq \frac{3}{2}, 0 \leq θ < 2 π

Für

cos (x) \geq a

, wenn

- 1 < a < 1

dann

- arccos (a) + 2 πn \leq x \leq arccos (a) + 2 πn

- arccos (\frac{3}{2}) + 2 πn \leq θ \leq arccos (\frac{3}{2}) + 2 πn

Vereinfache

- arccos (\frac{3}{2}) : - \frac{π}{6}

- arccos (\frac{3}{2})

Verwende die folgende triviale Identität:

arccos (\frac{3}{2}) = \frac{π}{6}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= - \frac{π}{6}

Vereinfache

arccos (\frac{3}{2}) : \frac{π}{6}

arccos (\frac{3}{2})

Verwende die folgende triviale Identität:

arccos (\frac{3}{2}) = \frac{π}{6}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{6}

- \frac{π}{6} + 2 πn \leq θ \leq \frac{π}{6} + 2 πn

Kombiniere die Bereiche

- \frac{π}{6} + 2 πn \leq θ \leq \frac{π}{6} + 2 πn and 0 \leq θ < 2 π

0 \leq θ \leq \frac{π}{6} or 2 π - \frac{π}{6} \leq θ < 2 π

Graph

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