tan(x)>= 1,0<= x<= 2pi

Lösung

tan (x) \geq 1, 0 \leq x \leq 2 π

\frac{π}{4} \leq x < \frac{π}{2} or π + \frac{π}{4} \leq x < π + \frac{π}{2}

Intervall-Notation

[\frac{π}{4}, \frac{π}{2}) \cup [π + \frac{π}{4}, π + \frac{π}{2})

Dezimale

0.78539 \dots \leq x < 1.57079 \dots or 3.92699 \dots \leq x < 4.71238 \dots

Schritte zur Lösung

tan (x) \geq 1, 0 \leq x \leq 2 π

Wenn

tan (x) \geq a

dann

arctan (a) + πn \leq x < \frac{π}{2} + πn

arctan (1) + πn \leq x < \frac{π}{2} + πn

Vereinfache

arctan (1) : \frac{π}{4}

arctan (1)

Verwende die folgende triviale Identität:

arctan (1) = \frac{π}{4}

x 0 \frac{3}{3} 13 arctan (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} arctan (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ}

= \frac{π}{4}

\frac{π}{4} + πn \leq x < \frac{π}{2} + πn

Kombiniere die Bereiche

\frac{π}{4} + πn \leq x < \frac{π}{2} + πn and 0 \leq x \leq 2 π

\frac{π}{4} \leq x < \frac{π}{2} or π + \frac{π}{4} \leq x < π + \frac{π}{2}

1 + sec (x) \geq 0

- sin (x) \leq 0.5

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

cos^{2} (x) < \frac{2}{2} + sin^{2} (x)

tan (x) < - 1.5