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Popular Trigonometria >

sin(θ)= 9/41 \land cos(θ)>0

Solução

sin (θ) = \frac{9}{41} and cos (θ) > 0

Solução

θ = 0.22131 \dots + 2 πn

+1

Notação de intervalo

θ = 0.22131 \dots + 2 πn

Passos da solução

sin (θ) = \frac{9}{41} and cos (θ) > 0

sin (θ) = \frac{9}{41} : θ = 0.22131 \dots + 2 πn, θ = π - 0.22131 \dots + 2 πn

sin (θ) = \frac{9}{41}

Aplique as propriedades trigonométricas inversas

sin (θ) = \frac{9}{41}

Soluções gerais para

sin (θ) = \frac{9}{41}

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

θ = arcsin (\frac{9}{41}) + 2 πn, θ = π - arcsin (\frac{9}{41}) + 2 πn

θ = arcsin (\frac{9}{41}) + 2 πn, θ = π - arcsin (\frac{9}{41}) + 2 πn

Mostrar soluções na forma decimal

θ = 0.22131 \dots + 2 πn, θ = π - 0.22131 \dots + 2 πn

cos (θ) > 0 : - \frac{π}{2} + 2 πn < θ < \frac{π}{2} + 2 πn

cos (θ) > 0

Para

cos (x) > a

, se

- 1 \leq a < 1

então

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

- arccos (0) + 2 πn < θ < 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

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}

- \frac{π}{2} + 2 πn < θ < \frac{π}{2} + 2 πn

Combinar os intervalos

(θ = 0.22131 \dots + 2 πn or θ = π - 0.22131 \dots + 2 πn) and - \frac{π}{2} + 2 πn < θ < \frac{π}{2} + 2 πn

Junte intervalos que se sobrepoem

θ = 0.22131 \dots + 2 πn

Exemplos populares

derivada de arcsech(cos(5x)0)<x< pi/5

\frac{d}{d x} (arcsech (cos (5 x)) 0) < x < \frac{π}{5}

tan(θ)>0\land cos(θ)<0

tan (θ) > 0 and cos (θ) < 0

(11pi)/2 <= arctan(θ)<= (13pi)/9

\frac{11 π}{2} \leq arctan (θ) \leq \frac{13 π}{9}

0<= tan^2(x)<= 3

0 \leq tan^{2} (x) \leq 3

tan(θ)=2\land cos(θ)<0

tan (θ) = 2 and cos (θ) < 0