tan(x)>= 3

Lösung

tan (x) \geq 3

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

Intervall-Notation

[arctan (3) + πn, \frac{π}{2} + πn)

Dezimale

1.24904 \dots + πn \leq x < 1.57079 \dots + πn

Schritte zur Lösung

tan (x) \geq 3

Wenn

tan (x) \geq a

dann

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

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

2 cos^{2} (x) < \frac{3}{2}

sin (3 x) cos (3 x) - \frac{1}{4} \geq 0

sin (x) + 3 cos (x) > 0

tan (x) < - 1

sin (x) < \frac{3}{2}