cot(x)<1

Lösung

cot (x) < 1

\frac{π}{4} + πn < x < π + πn

Intervall-Notation

(\frac{π}{4} + πn, π + πn)

Dezimale

0.78539 \dots + πn < x < 3.14159 \dots + πn

Schritte zur Lösung

cot (x) < 1

Wenn

cot (x) < a

dann

arccot (a) + πn < x < π + πn

arccot (1) + πn < x < π + πn

Vereinfache

arccot (1) : \frac{π}{4}

arccot (1)

Verwende die folgende triviale Identität:

arccot (1) = \frac{π}{4}

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

= \frac{π}{4}

\frac{π}{4} + πn < x < π + πn

2 sin (x) - 1 \geq 0

sin^{2} (x) + \frac{1}{2} \cdot sin (x) > 0

- sin (x) > 0

arctan (x) \leq 1

\frac{2 sin ( x ) + 1}{2 cos ( x )} \leq 0