cot(x)>= 1

Lösung

cot (x) \geq 1

πn < x \leq \frac{π}{4} + πn

Intervall-Notation

(πn, \frac{π}{4} + πn]

Dezimale

πn < x \leq 0.78539 \dots + πn

Schritte zur Lösung

cot (x) \geq 1

Wenn

cot (x) \geq a

dann

πn < x \leq arccot (a) + πn

πn < x \leq arccot (1) + 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}

πn < x \leq \frac{π}{4} + πn

- (- 1 - cos (t)) < 0

\frac{4 cos ( x ) + 3}{3 cos ( x ) + 1} < 2

sin^{2} (2 x) \leq \frac{1}{2}

\frac{1 - 2 cos ^{2} ( x )}{tan ( x )} > 0

2 cos^{2} (x) + cos (x) > 0