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cot^2(x)>= 1/3

Solución

cot^{2} (x) \geq \frac{1}{3}

Solución

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

+2

Notación de intervalos

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

Notación decimal

πn < x \leq 1.04719 \dots + πn or 2.09439 \dots + πn \leq x < 3.14159 \dots + πn

Pasos de solución

cot^{2} (x) \geq \frac{1}{3}

Para

u^{n} \geq a

, si

n

es par entonces

cot (x) \leq - \frac{1}{3} or cot (x) \geq \frac{1}{3}

cot (x) \leq - \frac{1}{3} : \frac{2 π}{3} + πn \leq x < π + πn

cot (x) \leq - \frac{1}{3}

Pi

cot (x) \leq a

entonces

arccot (a) + πn \leq x < π + πn

arccot (- \frac{1}{3}) + πn \leq x < π + πn

Simplificar

arccot (- \frac{1}{3}) : \frac{2 π}{3}

arccot (- \frac{1}{3})

Utilizar la siguiente identidad trivial:

arccot (- \frac{1}{3}) = \frac{2 π}{3}

x - 3 - 1 - \frac{3}{3} 0 \frac{3}{3} 13 arccot (x) \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} arccot (x) 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ}

= \frac{2 π}{3}

\frac{2 π}{3} + πn \leq x < π + πn

cot (x) \geq \frac{1}{3} : πn < x \leq \frac{π}{3} + πn

cot (x) \geq \frac{1}{3}

Pi

cot (x) \geq a

entonces

πn < x \leq arccot (a) + πn

πn < x \leq arccot (\frac{1}{3}) + πn

Simplificar

arccot (\frac{1}{3}) : \frac{π}{3}

arccot (\frac{1}{3})

Utilizar la siguiente identidad trivial:

arccot (\frac{1}{3}) = \frac{π}{3}

x - 3 - 1 - \frac{3}{3} 0 \frac{3}{3} 13 arccot (x) \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} arccot (x) 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ}

= \frac{π}{3}

πn < x \leq \frac{π}{3} + πn

Combinar los rangos

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

Mezclar intervalos sobrepuestos

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

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