English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor x,e^{cos(x)}(1-cos(y))=c

Solución

resolver para

x, e^{c o s (x)} (1 - cos (y)) = c

x = arccos (ln (\frac{c}{1 - cos ( y )})) + 2 πn, x = - arccos (ln (\frac{c}{1 - cos ( y )})) + 2 πn

Pasos de solución

e^{c o s (x)} (1 - cos (y)) = c

Dividir ambos lados entre

1 - cos (y)

e^{c o s (x)} (1 - cos (y)) = c

Dividir ambos lados entre

1 - cos (y)

\frac{e ^{c o s (x)} ( 1 - cos ( y ) )}{1 - cos ( y )} = \frac{c}{1 - cos ( y )}

Simplificar

e^{c o s (x)} = \frac{c}{1 - cos ( y )}

e^{c o s (x)} = \frac{c}{1 - cos ( y )}

Aplicar las leyes de los exponentes

e^{c o s (x)} = \frac{c}{1 - cos ( y )}

f (x) = g (x)

, entonces

ln (f (x)) = ln (g (x))

ln (e^{c o s (x)}) = ln (\frac{c}{1 - cos ( y )})

Aplicar las propiedades de los logaritmos:

ln (e^{a}) = a

ln (e^{c o s (x)}) = cos (x)

cos (x) = ln (\frac{c}{1 - cos ( y )})

cos (x) = ln (\frac{c}{1 - cos ( y )})

Aplicar propiedades trigonométricas inversas

cos (x) = ln (\frac{c}{1 - cos ( y )})

Soluciones generales para

cos (x) = ln (\frac{c}{1 - cos ( y )})

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

x = arccos (ln (\frac{c}{1 - cos ( y )})) + 2 πn, x = - arccos (ln (\frac{c}{1 - cos ( y )})) + 2 πn

x = arccos (ln (\frac{c}{1 - cos ( y )})) + 2 πn, x = - arccos (ln (\frac{c}{1 - cos ( y )})) + 2 πn

22 = 2 sec (θ)

cos^{4} (x) - sin^{4} (x) = \frac{1}{2}

sin (x) + cos (x) = tan (x) + 1

cos (x) = \frac{1}{4}, 0 \leq x \leq 2 π

tan^{2} (x) + 1 = a^{2}