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cos(θ)= 1/(sqrt(1^2+(4)^2+8^2))

Solución

cos (θ) = \frac{1}{1 ^{2} + ( 4 ) ^{2} + 8 ^{2}}

θ = 1.45945 \dots + 2 πn, θ = 2 π - 1.45945 \dots + 2 πn

radianes

θ = 1.45945 \dots + 6.28318 \dots n, θ = 4.82372 \dots + 6.28318 \dots n

Pasos de solución

cos (θ) = \frac{1}{1 ^{2} + ( 4 ) ^{2} + 8 ^{2}}

Simplificar

\frac{1}{1 ^{2} + 4 ^{2} + 8 ^{2}} : \frac{1}{9}

\frac{1}{1 ^{2} + 4 ^{2} + 8 ^{2}}

Aplicar la regla

1^{a} = 1

1^{2} = 1

= \frac{1}{1 + 4 ^{2} + 8 ^{2}}

1 + 4^{2} + 8^{2} = 81

1 + 4^{2} + 8^{2}

4^{2} = 16

= 1 + 16 + 8^{2}

8^{2} = 64

= 1 + 16 + 64

Sumar:

1 + 16 + 64 = 81

= 81

= \frac{1}{81}

81 = 9

81

Descomponer el número en factores primos:

81 = 9^{2}

= 9^{2}

Aplicar las leyes de los exponentes:

9^{2} = 9

= 9

= \frac{1}{9}

Aplicar propiedades trigonométricas inversas

cos (θ) = \frac{1}{9}

Soluciones generales para

cos (θ) = \frac{1}{9}

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

θ = arccos (\frac{1}{9}) + 2 πn, θ = 2 π - arccos (\frac{1}{9}) + 2 πn

θ = arccos (\frac{1}{9}) + 2 πn, θ = 2 π - arccos (\frac{1}{9}) + 2 πn

Mostrar soluciones en forma decimal

θ = 1.45945 \dots + 2 πn, θ = 2 π - 1.45945 \dots + 2 πn

2 + 3 cot^{2} (x) - 3 csc (x) = 5

- 2 = - 2 + tan (θ)

cos (A) = 0.65

sin (3 x - \frac{π}{4}) = \frac{2}{2}

sin (2 θ) = \frac{- 3}{4}, \frac{3 π}{2} \leq 2 θ \leq 2 π