tan(x)>sqrt(3),0<= x<= 2pi

Lösung

tan (x) > 3, 0 \leq x \leq 2 π

\frac{π}{3} < x < \frac{π}{2} or π + \frac{π}{3} < x < π + \frac{π}{2}

Intervall-Notation

(\frac{π}{3}, \frac{π}{2}) \cup (π + \frac{π}{3}, π + \frac{π}{2})

Dezimale

1.04719 \dots < x < 1.57079 \dots or 4.18879 \dots < x < 4.71238 \dots

Schritte zur Lösung

tan (x) > 3, 0 \leq x \leq 2 π

Wenn

tan (x) > a

dann

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

arctan (3) + πn < x < \frac{π}{2} + πn

Vereinfache

arctan (3) : \frac{π}{3}

arctan (3)

Verwende die folgende triviale Identität:

arctan (3) = \frac{π}{3}

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}{3}

\frac{π}{3} + πn < x < \frac{π}{2} + πn

Kombiniere die Bereiche

\frac{π}{3} + πn < x < \frac{π}{2} + πn and 0 \leq x \leq 2 π

\frac{π}{3} < x < \frac{π}{2} or π + \frac{π}{3} < x < π + \frac{π}{2}

- 1 \leq \frac{2 + sin ( x )}{3}

sin (x) > \frac{2}{2}, 0 \leq x \leq 2 π

sin (x) < \frac{1}{4}

2 sin^{2} (x) - sin (x) - 1 < 0

cos^{2} (\frac{x}{3}) \leq sin^{2} (\frac{x}{3}) - \frac{1}{2}