cos(x)>= (sqrt(3))/2

Lösung

cos (x) \geq \frac{3}{2}

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

Intervall-Notation

[- \frac{π}{6} + 2 πn, \frac{π}{6} + 2 πn]

Dezimale

- 0.52359 \dots + 2 πn \leq x \leq 0.52359 \dots + 2 πn

Schritte zur Lösung

cos (x) \geq \frac{3}{2}

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 x \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 x \leq \frac{π}{6} + 2 πn

cos (x) - 1 \geq 0

2 sin (\frac{x}{2}) - 1 > 0

sin (3 x) < \frac{2}{2}

sin (2 x - \frac{π}{12}) \leq \frac{2}{2}

\frac{π}{3} < tan (t), - csc (2 t)