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Popular Trigonometria >

(cos(x))^2>= 1

Solução

(cos (x))^{2} \geq 1

Solução

x = π + 2 πn

+1

Decimal

x = 3.14159 \dots + 2 πn

Passos da solução

(cos (x))^{2} \geq 1

Para

u^{n} \geq a

, se

n

é par então

u \leq - n a or u \geq n a

cos (x) \leq - 1 or cos (x) \geq 1

cos (x) \leq - 1 : x = π + 2 πn

cos (x) \leq - 1

Para

cos (x) \leq a

, se

- 1 < a < 1

então

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

arccos (- 1) + 2 πn \leq x \leq 2 π - arccos (- 1) + 2 πn

Simplificar

arccos (- 1) : π

arccos (- 1)

Utilizar a seguinte identidade 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

2 π - arccos (- 1) : π

2 π - arccos (- 1)

Utilizar a seguinte identidade 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}

= 2 π - π

Somar elementos similares:

2 π - π = π

= π

π + 2 πn \leq x \leq π + 2 πn

Simplificar

x = π + 2 πn

cos (x) \geq 1 :

Sem solução para

x \in R

cos (x) \geq 1

Para

cos (x) \geq a

, se

- 1 < a < 1

então

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

- arccos (1) + 2 πn \leq x \leq arccos (1) + 2 πn

Simplificar

- arccos (1) : 0

- arccos (1)

Utilizar a seguinte identidade 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

= 0

Simplificar

arccos (1) : 0

arccos (1)

Utilizar a seguinte identidade 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

0 + 2 πn \leq x \leq 0 + 2 πn

Simplificar

Sem solu \overset{c}{\c} \tilde{a} o para x \in R

Combinar os intervalos

x = π + 2 πn or Falso para todo x \in R

Junte intervalos que se sobrepoem

x = π + 2 πn

Exemplos populares

sin(x)<= (sqrt(3))/2 ,-pi<= x<= pi

sin (x) \leq \frac{3}{2}, - π \leq x \leq π

2cos(4x-pi/3)-1<= 0

2 cos (4 x - \frac{π}{3}) - 1 \leq 0

sin (x) \leq - \frac{5}{2}

arccos((x-5)/(10))-pi/3 >= 0

arccos (\frac{x - 5}{10}) - \frac{π}{3} \geq 0

tan(3x-1)>-sqrt(3)

tan (3 x - 1) > - 3