What is the general solution for solvefor y,2e^x-sin(y)=x ?

The general solution for solvefor y,2e^x-sin(y)=x is y=arcsin(-x+2e^x)+2pin,y=pi+arcsin(x-2e^x)+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor y,2e^x-sin(y)=x

solve for

y, 2 e^{x} - sin (y) = x

y = arcsin (- x + 2 e^{x}) + 2 πn, y = π + arcsin (x - 2 e^{x}) + 2 πn

Solution steps

2 e^{x} - sin (y) = x

Move

2 e^{x}

to the right side

2 e^{x} - sin (y) = x

Subtract

2 e^{x}

from both sides

2 e^{x} - sin (y) - 2 e^{x} = x - 2 e^{x}

Simplify

- sin (y) = x - 2 e^{x}

- sin (y) = x - 2 e^{x}

Divide both sides by

- 1

- sin (y) = x - 2 e^{x}

Divide both sides by

- 1

\frac{- sin ( y )}{- 1} = \frac{x}{- 1} - \frac{2 e ^{x}}{- 1}

Simplify

\frac{- sin ( y )}{- 1} = \frac{x}{- 1} - \frac{2 e ^{x}}{- 1}

Simplify

\frac{- sin ( y )}{- 1} : sin (y)

\frac{- sin ( y )}{- 1}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{sin ( y )}{1}

Apply rule

\frac{a}{1} = a

= sin (y)

Simplify

\frac{x}{- 1} - \frac{2 e ^{x}}{- 1} : - x + 2 e^{x}

\frac{x}{- 1} - \frac{2 e ^{x}}{- 1}

Apply rule

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

= \frac{x - 2 e ^{x}}{- 1}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{x - 2 e ^{x}}{1}

Apply rule

\frac{a}{1} = a

= - (x - 2 e^{x})

Distribute parentheses

= - x - (- 2 e^{x})

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - x + 2 e^{x}

sin (y) = - x + 2 e^{x}

sin (y) = - x + 2 e^{x}

sin (y) = - x + 2 e^{x}

Apply trig inverse properties

sin (y) = - x + 2 e^{x}

General solutions for

sin (y) = - x + 2 e^{x}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

y = arcsin (- x + 2 e^{x}) + 2 πn, y = π + arcsin (x - 2 e^{x}) + 2 πn

y = arcsin (- x + 2 e^{x}) + 2 πn, y = π + arcsin (x - 2 e^{x}) + 2 πn

sec^{2} (x) = \frac{1}{2}

3 sin^{2} (x) = 3 cos^{2} (x)

9 tan (\frac{x}{2}) + 7 = 5 tan (\frac{x}{2}) + 3

3 - 3 cos (θ) = 3

\frac{1 - tanh ( 2 x )}{1 + tanh ( 2 x )} = 2

What is the general solution for solvefor y,2e^x-sin(y)=x ?
The general solution for solvefor y,2e^x-sin(y)=x is y=arcsin(-x+2e^x)+2pin,y=pi+arcsin(x-2e^x)+2pin