es

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

-sqrt(3)+3=2sqrt(6)cos(θ)

Solución

- 3 + 3 = 26 cos (θ)

Solución

θ = 1.30899 \dots + 2 πn, θ = 2 π - 1.30899 \dots + 2 πn

+1

Grados

θ = 7 5^{\circ} + 36 0^{\circ} n, θ = 28 5^{\circ} + 36 0^{\circ} n

Pasos de solución

- 3 + 3 = 26 cos (θ)

Intercambiar lados

26 cos (θ) = - 3 + 3

Dividir ambos lados entre

26

26 cos (θ) = - 3 + 3

Dividir ambos lados entre

26

\frac{2 6 cos ( θ )}{2 6} = - \frac{3}{2 6} + \frac{3}{2 6}

Simplificar

\frac{2 6 cos ( θ )}{2 6} = - \frac{3}{2 6} + \frac{3}{2 6}

Simplificar

\frac{2 6 cos ( θ )}{2 6} : cos (θ)

\frac{2 6 cos ( θ )}{2 6}

Dividir:

\frac{2}{2} = 1

= \frac{6 cos ( θ )}{6}

Eliminar los terminos comunes:

6

= cos (θ)

Simplificar

- \frac{3}{2 6} + \frac{3}{2 6} : \frac{2 ( - 1 + 3 )}{4}

- \frac{3}{2 6} + \frac{3}{2 6}

Aplicar la regla

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- 3 + 3}{2 6}

Factorizar

- 3 + 3 : 3 (- 1 + 3)

- 3 + 3

3 = 33

= - 3 + 33

Factorizar el termino común

3

= 3 (- 1 + 3)

= \frac{3 ( - 1 + 3 )}{2 6}

Factorizar

6 : 23

Factorizar

6 = 2 \cdot 3

= 2 \cdot 3

Aplicar las leyes de los exponentes:

= 23

= \frac{3 ( 3 - 1 )}{2 2 3}

Eliminar los terminos comunes:

3

= \frac{- 1 + 3}{2 2}

Racionalizar

\frac{- 1 + 3}{2 2} : \frac{2 ( 3 - 1 )}{2 ^{2}}

\frac{- 1 + 3}{2 2}

Multiplicar por el conjugado

\frac{2}{2}

= \frac{( - 1 + 3 ) 2}{2 2 2}

222 = 2^{2}

222

Aplicar las leyes de los exponentes:

a^{b} \cdot a^{c} = a^{b + c}

222 = 2 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2} + \frac{1}{2}}

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

Sumar elementos similares:

\frac{1}{2} + \frac{1}{2} = 2 \cdot \frac{1}{2}

= 2^{1 + 2 \cdot \frac{1}{2}}

2 \cdot \frac{1}{2} = 1

2 \cdot \frac{1}{2}

Multiplicar fracciones:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

Eliminar los terminos comunes:

2

= 1

= 2^{1 + 1}

Sumar:

1 + 1 = 2

= 2^{2}

= \frac{2 ( - 1 + 3 )}{2 ^{2}}

= \frac{2 ( 3 - 1 )}{2 ^{2}}

2^{2} = 4

= \frac{2 ( 3 - 1 )}{4}

cos (θ) = \frac{2 ( - 1 + 3 )}{4}

cos (θ) = \frac{2 ( - 1 + 3 )}{4}

cos (θ) = \frac{2 ( - 1 + 3 )}{4}

Aplicar propiedades trigonométricas inversas

cos (θ) = \frac{2 ( - 1 + 3 )}{4}

Soluciones generales para

cos (θ) = \frac{2 ( - 1 + 3 )}{4}

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

θ = arccos (\frac{2 ( - 1 + 3 )}{4}) + 2 πn, θ = 2 π - arccos (\frac{2 ( - 1 + 3 )}{4}) + 2 πn

θ = arccos (\frac{2 ( - 1 + 3 )}{4}) + 2 πn, θ = 2 π - arccos (\frac{2 ( - 1 + 3 )}{4}) + 2 πn

Mostrar soluciones en forma decimal

θ = 1.30899 \dots + 2 πn, θ = 2 π - 1.30899 \dots + 2 πn

Ejemplos populares

sin(pi/2-x)=-1,0<= x<= 360

sin (\frac{π}{2} - x) = - 1, 0^{\circ} \leq x \leq 36 0^{\circ}

2 sin (5 x) = - 1

4sin^2(x)+8cos(x)-7=0

4 sin^{2} (x) + 8 cos (x) - 7 = 0

6(cos(b))^2+cos(b)-1=0

6 (cos (b))^{2} + cos (b) - 1 = 0

sin(θ)=(1.33)/(2.42)

sin (θ) = \frac{1.33}{2.42}