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cos^2(x)<1

Solución

cos^{2} (x) < 1

Solución

2 πn < x < π + 2 πn or π + 2 πn < x < 2 π + 2 πn

+2

Notación de intervalos

(2 πn, π + 2 πn) \cup (π + 2 πn, 2 π + 2 πn)

Notación decimal

2 πn < x < 3.14159 \dots + 2 πn or 3.14159 \dots + 2 πn < x < 6.28318 \dots + 2 πn

Pasos de solución

cos^{2} (x) < 1

Para

u^{n} < a

, si

n

es par entonces

- 1 < cos (x) < 1

Pi

a < u < b

entonces

a < u and u < b

- 1 < cos (x) and cos (x) < 1

- 1 < cos (x) : - π + 2 πn < x < π + 2 πn

- 1 < cos (x)

Intercambiar lados

cos (x) > - 1

Para

cos (x) > a

, si

- 1 \leq a < 1

entonces

- arccos (a) + 2 πn < x < arccos (a) + 2 πn

- arccos (- 1) + 2 πn < x < arccos (- 1) + 2 πn

Simplificar

- arccos (- 1) : - π

- arccos (- 1)

Utilizar la siguiente identidad trivial:

arccos (- 1) = π

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= - π

Simplificar

arccos (- 1) : π

arccos (- 1)

Utilizar la siguiente identidad trivial:

arccos (- 1) = π

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= π

- π + 2 πn < x < π + 2 πn

cos (x) < 1 : 2 πn < x < 2 π + 2 πn

cos (x) < 1

Para

cos (x) < a

, si

- 1 < a \leq 1

entonces

arccos (a) + 2 πn < x < 2 π - arccos (a) + 2 πn

arccos (1) + 2 πn < x < 2 π - arccos (1) + 2 πn

Simplificar

arccos (1) : 0

arccos (1)

Utilizar la siguiente identidad trivial:

arccos (1) = 0

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= 0

Simplificar

2 π - arccos (1) : 2 π

2 π - arccos (1)

Utilizar la siguiente identidad trivial:

arccos (1) = 0

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= 2 π - 0

2 π - 0 = 2 π

= 2 π

0 + 2 πn < x < 2 π + 2 πn

Simplificar

2 πn < x < 2 π + 2 πn

Combinar los rangos

- π + 2 πn < x < π + 2 πn and 2 πn < x < 2 π + 2 πn

Mezclar intervalos sobrepuestos

2 πn < x < π + 2 πn or π + 2 πn < x < 2 π + 2 πn

Ejemplos populares

sec^{2} (x) \leq \frac{4}{3}

cos(x)>(sqrt(3))/2

cos (x) > \frac{3}{2}

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

sin (x) \leq 1

tan (x) \geq 0