English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solución

cos (2 x) \geq - \frac{3}{2}

Solución

- \frac{5 π}{12} + πn \leq x \leq \frac{5 π}{12} + πn

Notación de intervalos

[- \frac{5 π}{12} + πn, \frac{5 π}{12} + πn]

Notación decimal

- 1.30899 \dots + πn \leq x \leq 1.30899 \dots + πn

Pasos de solución

cos (2 x) \geq - \frac{3}{2}

Para

cos (x) \geq a

, si

- 1 < a < 1

entonces

- arccos (a) + 2 πn \leq x \leq arccos (a) + 2 πn

- arccos (- \frac{3}{2}) + 2 πn \leq 2 x \leq arccos (- \frac{3}{2}) + 2 πn

a \leq u \leq b

entonces

a \leq u and u \leq b

- arccos (- \frac{3}{2}) + 2 πn \leq 2 x and 2 x \leq arccos (- \frac{3}{2}) + 2 πn

- arccos (- \frac{3}{2}) + 2 πn \leq 2 x : x \geq - \frac{5 π}{12} + πn

- arccos (- \frac{3}{2}) + 2 πn \leq 2 x

Intercambiar lados

2 x \geq - arccos (- \frac{3}{2}) + 2 πn

Simplificar

- arccos (- \frac{3}{2}) + 2 πn : - \frac{5 π}{6} + 2 πn

- arccos (- \frac{3}{2}) + 2 πn

Utilizar la siguiente identidad trivial:

arccos (- \frac{3}{2}) = \frac{5 π}{6}

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}

= - \frac{5 π}{6} + 2 πn

2 x \geq - \frac{5 π}{6} + 2 πn

Dividir ambos lados entre

2

2 x \geq - \frac{5 π}{6} + 2 πn

Dividir ambos lados entre

2

\frac{2 x}{2} \geq - \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

Simplificar

\frac{2 x}{2} \geq - \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

Simplificar

\frac{2 x}{2} : x

\frac{2 x}{2}

Dividir:

\frac{2}{2} = 1

= x

Simplificar

- \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2} : - \frac{5 π}{12} + πn

- \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{6}}{2} = \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2}

Aplicar las propiedades de las fracciones:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{5 π}{6 \cdot 2}

Multiplicar los numeros:

6 \cdot 2 = 12

= \frac{5 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Dividir:

\frac{2}{2} = 1

= πn

= - \frac{5 π}{12} + πn

x \geq - \frac{5 π}{12} + πn

x \geq - \frac{5 π}{12} + πn

x \geq - \frac{5 π}{12} + πn

2 x \leq arccos (- \frac{3}{2}) + 2 πn : x \leq \frac{5 π}{12} + πn

2 x \leq arccos (- \frac{3}{2}) + 2 πn

Simplificar

arccos (- \frac{3}{2}) + 2 πn : \frac{5 π}{6} + 2 πn

arccos (- \frac{3}{2}) + 2 πn

Utilizar la siguiente identidad trivial:

arccos (- \frac{3}{2}) = \frac{5 π}{6}

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}

= \frac{5 π}{6} + 2 πn

2 x \leq \frac{5 π}{6} + 2 πn

Dividir ambos lados entre

2

2 x \leq \frac{5 π}{6} + 2 πn

Dividir ambos lados entre

2

\frac{2 x}{2} \leq \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

Simplificar

\frac{2 x}{2} \leq \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

Simplificar

\frac{2 x}{2} : x

\frac{2 x}{2}

Dividir:

\frac{2}{2} = 1

= x

Simplificar

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2} : \frac{5 π}{12} + πn

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{6}}{2} = \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2}

Aplicar las propiedades de las fracciones:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{5 π}{6 \cdot 2}

Multiplicar los numeros:

6 \cdot 2 = 12

= \frac{5 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Dividir:

\frac{2}{2} = 1

= πn

= \frac{5 π}{12} + πn

x \leq \frac{5 π}{12} + πn

x \leq \frac{5 π}{12} + πn

x \leq \frac{5 π}{12} + πn

Combinar los rangos

x \geq - \frac{5 π}{12} + πn and x \leq \frac{5 π}{12} + πn

Mezclar intervalos sobrepuestos

- \frac{5 π}{12} + πn \leq x \leq \frac{5 π}{12} + πn

Ejemplos populares

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

2 tan (x) < 2, 0 \leq x \leq 2 π

cot(x)>sqrt(3)

cot (x) > 3

cos^2(x)> 1/4

cos^{2} (x) > \frac{1}{4}

cos^2(x)>= 0

cos^{2} (x) \geq 0

cot(x)>=-sqrt(3)

cot (x) \geq - 3