es

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos(x)>= (-1)/2

Solución

cos (x) \geq \frac{- 1}{2}

Solución

- \frac{2 π}{3} + 2 πn \leq x \leq \frac{2 π}{3} + 2 πn

+2

Notación de intervalos

[- \frac{2 π}{3} + 2 πn, \frac{2 π}{3} + 2 πn]

Notación decimal

- 2.09439 \dots + 2 πn \leq x \leq 2.09439 \dots + 2 πn

Pasos de solución

cos (x) \geq \frac{- 1}{2}

Aplicar las propiedades de las fracciones:

\frac{- a}{b} = - \frac{a}{b}

cos (x) \geq - \frac{1}{2}

Para

cos (x) \geq a

, si

- 1 < a < 1

entonces

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

- arccos (- \frac{1}{2}) + 2 πn \leq x \leq arccos (- \frac{1}{2}) + 2 πn

Simplificar

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

- arccos (- \frac{1}{2})

Utilizar la siguiente identidad trivial:

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

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{2 π}{3}

Simplificar

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

arccos (- \frac{1}{2})

Utilizar la siguiente identidad trivial:

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

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{2 π}{3}

- \frac{2 π}{3} + 2 πn \leq x \leq \frac{2 π}{3} + 2 πn

Ejemplos populares

(cos^2(x))/(1-cos(x))<0

\frac{cos ^{2} ( x )}{1 - cos ( x )} < 0

pi/2-arctan(n)<= 10^{-3}

\frac{π}{2} - arctan (n) \leq 1 0^{- 3}

cos(3x)<= (sqrt(3))/2

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

cos (2 x) < \frac{1}{2}

sin (x) \geq - 1