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sin(x)= 1/(sqrt(5))\land cos(x)<0

Solução

sin (x) = \frac{1}{5} and cos (x) < 0

x = π - 0.46364 \dots + 2 πn

Decimal

x = 2.67794 \dots + 2 πn

Passos da solução

sin (x) = \frac{1}{5} and cos (x) < 0

sin (x) = \frac{1}{5} : x = 0.46364 \dots + 2 πn, x = π - 0.46364 \dots + 2 πn

sin (x) = \frac{1}{5}

Simplificar

\frac{1}{5} : \frac{5}{5}

\frac{1}{5}

Multiplicar pelo conjugado

\frac{5}{5}

= \frac{1 \cdot 5}{5 5}

1 \cdot 5 = 5

55 = 5

55

Aplicar as propriedades dos radicais:

a a = a

55 = 5

= 5

= \frac{5}{5}

sin (x) = \frac{5}{5}

Aplique as propriedades trigonométricas inversas

sin (x) = \frac{5}{5}

Soluções gerais para

sin (x) = \frac{5}{5}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{5}{5}) + 2 πn, x = π - arcsin (\frac{5}{5}) + 2 πn

x = arcsin (\frac{5}{5}) + 2 πn, x = π - arcsin (\frac{5}{5}) + 2 πn

Mostrar soluções na forma decimal

x = 0.46364 \dots + 2 πn, x = π - 0.46364 \dots + 2 πn

cos (x) < 0 : \frac{π}{2} + 2 πn < x < \frac{3 π}{2} + 2 πn

cos (x) < 0

Para

cos (x) < a

, se

- 1 < a \leq 1

então

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

arccos (0) + 2 πn < x < 2 π - arccos (0) + 2 πn

Simplificar

arccos (0) : \frac{π}{2}

arccos (0)

Utilizar a seguinte identidade trivial:

arccos (0) = \frac{π}{2}

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{π}{2}

Simplificar

2 π - arccos (0) : \frac{3 π}{2}

2 π - arccos (0)

Utilizar a seguinte identidade trivial:

arccos (0) = \frac{π}{2}

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 - \frac{π}{2}

Simplificar

2 π - \frac{π}{2}

Converter para fração:

2 π = \frac{2 π 2}{2}

= \frac{2 π 2}{2} - \frac{π}{2}

Já que os denominadores são iguais, combinar as frações:

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

= \frac{2 π 2 - π}{2}

2 π 2 - π = 3 π

2 π 2 - π

Multiplicar os números:

2 \cdot 2 = 4

= 4 π - π

Somar elementos similares:

4 π - π = 3 π

= 3 π

= \frac{3 π}{2}

= \frac{3 π}{2}

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

Combinar os intervalos

(x = 0.46364 \dots + 2 πn or x = π - 0.46364 \dots + 2 πn) and \frac{π}{2} + 2 πn < x < \frac{3 π}{2} + 2 πn

Junte intervalos que se sobrepoem

x = π - 0.46364 \dots + 2 πn

cot (θ) < 0 and sec (θ) > 0

6 sin (x) 0 \leq x \leq \frac{3 π}{2}

- 1 \leq - cos (2 x) \leq 1

0 \leq 2 sin (3 x) + 1 < 2 π

tan (- \frac{π}{4}) \leq tan (\frac{a}{2}) \leq tan (\frac{π}{4})