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solvefor θ,sin^2(θ)= 1/4

Solução

resolver para

θ, sin^{2} (θ) = \frac{1}{4}

Solução

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

+1

Graus

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n, θ = 21 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n

Passos da solução

sin^{2} (θ) = \frac{1}{4}

Usando o método de substituição

sin^{2} (θ) = \frac{1}{4}

Sea:

sin (θ) = u

u^{2} = \frac{1}{4}

u^{2} = \frac{1}{4} : u = \frac{1}{2}, u = - \frac{1}{2}

u^{2} = \frac{1}{4}

Para

x^{2} = f (a)

as soluções são

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

u = \frac{1}{4}, u = - \frac{1}{4}

\frac{1}{4} = \frac{1}{2}

\frac{1}{4}

Aplicar a seguinte propriedade dos radicais:

assumindo que

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Fatorar o número:

4 = 2^{2}

= 2^{2}

Aplicar as propriedades dos radicais:

2^{2} = 2

= 2

= \frac{1}{2}

Aplicar a regra

1 = 1

= \frac{1}{2}

- \frac{1}{4} = - \frac{1}{2}

- \frac{1}{4}

Simplificar

\frac{1}{4} : \frac{1}{2}

\frac{1}{4}

Aplicar a seguinte propriedade dos radicais:

assumindo que

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Fatorar o número:

4 = 2^{2}

= 2^{2}

Aplicar as propriedades dos radicais:

2^{2} = 2

= 2

= \frac{1}{2}

= - \frac{1}{2}

Aplicar a regra

1 = 1

= - \frac{1}{2}

u = \frac{1}{2}, u = - \frac{1}{2}

Substituir na equação

u = sin (θ)

sin (θ) = \frac{1}{2}, sin (θ) = - \frac{1}{2}

sin (θ) = \frac{1}{2}, sin (θ) = - \frac{1}{2}

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (θ) = \frac{1}{2}

Soluções gerais para

sin (θ) = \frac{1}{2}

sin (x)

tabela de periodicidade com ciclo de

2 πn

:

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{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

sin (θ) = - \frac{1}{2}

Soluções gerais para

sin (θ) = - \frac{1}{2}

sin (x)

tabela de periodicidade com ciclo de

2 πn

:

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{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Combinar toda as soluções

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

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