tan(θ)>1

Lösung

tan (θ) > 1

\frac{π}{4} + πn < θ < \frac{π}{2} + πn

Intervall-Notation

(\frac{π}{4} + πn, \frac{π}{2} + πn)

Dezimale

0.78539 \dots + πn < θ < 1.57079 \dots + πn

Schritte zur Lösung

tan (θ) > 1

Wenn

tan (x) > a

dann

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

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

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

8 - 9 sin (x) cos (x) > 20

cos^{2} (x) - 2 cos (x) + 1 \leq 0

cos (x) sin (x) < 0

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