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sec^2(θ)-1=0,0<= θ<= 2pi

Solución

sec^{2} (θ) - 1 = 0, 0 \leq θ \leq 2 π

Solución

θ = 0, θ = 2 π, θ = π

+1

Grados

θ = 0^{\circ}, θ = 36 0^{\circ}, θ = 18 0^{\circ}

Pasos de solución

sec^{2} (θ) - 1 = 0, 0 \leq θ \leq 2 π

Usando el método de sustitución

sec^{2} (θ) - 1 = 0

Sea:

sec (θ) = u

u^{2} - 1 = 0

u^{2} - 1 = 0 : u = 1, u = - 1

u^{2} - 1 = 0

Mover

1

al lado derecho

u^{2} - 1 = 0

Sumar

1

a ambos lados

u^{2} - 1 + 1 = 0 + 1

Simplificar

u^{2} = 1

u^{2} = 1

Para

x^{2} = f (a)

las soluciones son

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Aplicar la regla

1 = 1

= 1

- 1 = - 1

- 1

Aplicar la regla

1 = 1

= - 1

u = 1, u = - 1

Sustituir en la ecuación

u = sec (θ)

sec (θ) = 1, sec (θ) = - 1

sec (θ) = 1, sec (θ) = - 1

sec (θ) = 1, 0 \leq θ \leq 2 π : θ = 0, θ = 2 π

sec (θ) = 1, 0 \leq θ \leq 2 π

Soluciones generales para

sec (θ) = 1

sec (x)

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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

θ = 0 + 2 πn

θ = 0 + 2 πn

Resolver

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

Soluciones para el rango

0 \leq θ \leq 2 π

θ = 0, θ = 2 π

sec (θ) = - 1, 0 \leq θ \leq 2 π : θ = π

sec (θ) = - 1, 0 \leq θ \leq 2 π

Soluciones generales para

sec (θ) = - 1

sec (x)

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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

θ = π + 2 πn

θ = π + 2 πn

Soluciones para el rango

0 \leq θ \leq 2 π

θ = π

Combinar toda las soluciones

θ = 0, θ = 2 π, θ = π

Ejemplos populares

2cos^2(θ)-3cos(θ)=0,(3pi)/2 <θ<2pi

2 cos^{2} (θ) - 3 cos (θ) = 0, \frac{3 π}{2} < θ < 2 π

cos (θ) = \frac{9}{17}

-1=(tan(x)-(3/4))/(1+tan(x)(3/4))

- 1 = \frac{tan ( x ) - ( \frac{3}{4} )}{1 + tan ( x ) ( \frac{3}{4} )}

cos (x) = - \frac{15}{13}

arctan((20)/x)-arctan((400)/x)=-9.72

arctan (\frac{20}{x}) - arctan (\frac{400}{x}) = - 9.72