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

Lösung

tan (x) \leq 1, 0 \leq x \leq 2 π

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

Intervall-Notation

[0, \frac{π}{4}] \cup (π - \frac{π}{2}, π + \frac{π}{4}] \cup (2 π - \frac{π}{2}, 2 π]

Dezimale

0 \leq x \leq 0.78539 \dots or 1.57079 \dots < x \leq 3.92699 \dots or 4.71238 \dots < x \leq 6.28318 \dots

Schritte zur Lösung

tan (x) \leq 1, 0 \leq x \leq 2 π

Wenn

tan (x) \leq a

dann

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

- \frac{π}{2} + πn < x \leq arctan (1) + 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}{2} + πn < x \leq \frac{π}{4} + πn

Kombiniere die Bereiche

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

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

(sin (x)) < \frac{1}{2}

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

\frac{( 4 \cdot cos ^{2} ( x ) - 3 )}{( sin ( x ) + cos ( x ) + 5 )} < 0

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

sin (θ) \leq \frac{π}{6}