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Popular >

sin(θ)=-(sqrt(3))/2 \land cos(θ)>= 0

Solución

sin (θ) = - \frac{3}{2} and cos (θ) \geq 0

θ = \frac{5 π}{3} + 2 πn

Notación decimal

θ = 5.23598 \dots + 2 πn

Pasos de solución

sin (θ) = - \frac{3}{2} and cos (θ) \geq 0

sin (θ) = - \frac{3}{2} : θ = \frac{4 π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

sin (θ) = - \frac{3}{2}

Soluciones generales para

sin (θ) = - \frac{3}{2}

tabla de valores periódicos con

2 πn

intervalos:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = \frac{4 π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

θ = \frac{4 π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

cos (θ) \geq 0 : - \frac{π}{2} + 2 πn \leq θ \leq \frac{π}{2} + 2 πn

cos (θ) \geq 0

Para

cos (x) \geq a

, si

- 1 < a < 1

entonces

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

- arccos (0) + 2 πn \leq θ \leq arccos (0) + 2 πn

Simplificar

- arccos (0) : - \frac{π}{2}

- arccos (0)

Utilizar la siguiente identidad 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 la siguiente identidad 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 \leq θ \leq \frac{π}{2} + 2 πn

Combinar los rangos

(θ = \frac{4 π}{3} + 2 πn or θ = \frac{5 π}{3} + 2 πn) and - \frac{π}{2} + 2 πn \leq θ \leq \frac{π}{2} + 2 πn

Mezclar intervalos sobrepuestos

θ = \frac{5 π}{3} + 2 πn

- 1 \leq \frac{π ( cos ( x ) - sin ( x ) )}{4 2} \leq 1

0 \leq cos (θ) \leq 1

- 2 sin^{2} (x) 0 < x < 360

- 1 \leq tan (\frac{x}{2} - \frac{π}{3}) \leq 3

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