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cos^2(t)= 1/2

Solução

cos^{2} (t) = \frac{1}{2}

Solução

t = 0.78539 \dots + 2 πn, t = 2 π - 0.78539 \dots + 2 πn, t = 2.35619 \dots + 2 πn, t = - 2.35619 \dots + 2 πn

+1

Graus

t = 4 5^{\circ} + 36 0^{\circ} n, t = 31 5^{\circ} + 36 0^{\circ} n, t = 13 5^{\circ} + 36 0^{\circ} n, t = - 13 5^{\circ} + 36 0^{\circ} n

Passos da solução

cos^{2} (t) = \frac{1}{2}

Usando o método de substituição

cos^{2} (t) = \frac{1}{2}

Sea:

cos (t) = u

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

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

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

Para

x^{2} = f (a)

as soluções são

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

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

Substituir na equação

u = cos (t)

cos (t) = \frac{1}{2}, cos (t) = - \frac{1}{2}

cos (t) = \frac{1}{2}, cos (t) = - \frac{1}{2}

cos (t) = \frac{1}{2} : t = arccos (\frac{1}{2}) + 2 πn, t = 2 π - arccos (\frac{1}{2}) + 2 πn

cos (t) = \frac{1}{2}

Aplique as propriedades trigonométricas inversas

cos (t) = \frac{1}{2}

Soluções gerais para

cos (t) = \frac{1}{2}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

t = arccos (\frac{1}{2}) + 2 πn, t = 2 π - arccos (\frac{1}{2}) + 2 πn

t = arccos (\frac{1}{2}) + 2 πn, t = 2 π - arccos (\frac{1}{2}) + 2 πn

cos (t) = - \frac{1}{2} : t = arccos (- \frac{1}{2}) + 2 πn, t = - arccos (- \frac{1}{2}) + 2 πn

cos (t) = - \frac{1}{2}

Aplique as propriedades trigonométricas inversas

cos (t) = - \frac{1}{2}

Soluções gerais para

cos (t) = - \frac{1}{2}

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

t = arccos (- \frac{1}{2}) + 2 πn, t = - arccos (- \frac{1}{2}) + 2 πn

t = arccos (- \frac{1}{2}) + 2 πn, t = - arccos (- \frac{1}{2}) + 2 πn

Combinar toda as soluções

t = arccos (\frac{1}{2}) + 2 πn, t = 2 π - arccos (\frac{1}{2}) + 2 πn, t = arccos (- \frac{1}{2}) + 2 πn, t = - arccos (- \frac{1}{2}) + 2 πn

Mostrar soluções na forma decimal

t = 0.78539 \dots + 2 πn, t = 2 π - 0.78539 \dots + 2 πn, t = 2.35619 \dots + 2 πn, t = - 2.35619 \dots + 2 πn

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