en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor x,(D^2-3D+2)y=cos(e^{-x})

Solution

solve for

x, (D^{2} - 3 D + 2) y = cos (e^{- x})

Solution

x = - ln (arccos (y (D^{2} - 3 D + 2)) + 2 πn), x = - ln (- arccos (y (D^{2} - 3 D + 2)) + 2 πn)

Solution steps

(D^{2} - 3 D + 2) y = cos (e^{- x})

Switch sides

cos (e^{- x}) = (D^{2} - 3 D + 2) y

Apply trig inverse properties

cos (e^{- x}) = (D^{2} - 3 D + 2) y

General solutions for

cos (e^{- x}) = (D^{2} - 3 D + 2) y

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

e^{- x} = arccos ((D^{2} - 3 D + 2) y) + 2 πn, e^{- x} = - arccos ((D^{2} - 3 D + 2) y) + 2 πn

e^{- x} = arccos ((D^{2} - 3 D + 2) y) + 2 πn, e^{- x} = - arccos ((D^{2} - 3 D + 2) y) + 2 πn

Solve

e^{- x} = arccos ((D^{2} - 3 D + 2) y) + 2 πn : x = - ln (arccos (y (D^{2} - 3 D + 2)) + 2 πn)

e^{- x} = arccos ((D^{2} - 3 D + 2) y) + 2 πn

Apply exponent rules

e^{- x} = arccos ((D^{2} - 3 D + 2) y) + 2 πn

If

f (x) = g (x)

, then

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

ln (e^{- x}) = ln (arccos ((D^{2} - 3 D + 2) y) + 2 πn)

Apply log rule:

ln (e^{a}) = a

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

- x = ln (arccos ((D^{2} - 3 D + 2) y) + 2 πn)

- x = ln (arccos ((D^{2} - 3 D + 2) y) + 2 πn)

Solve

- x = ln (arccos ((D^{2} - 3 D + 2) y) + 2 πn) : x = - ln (arccos (y (D^{2} - 3 D + 2)) + 2 πn)

- x = ln (arccos ((D^{2} - 3 D + 2) y) + 2 πn)

Divide both sides by

- 1

- x = ln (arccos ((D^{2} - 3 D + 2) y) + 2 πn)

Divide both sides by

- 1

\frac{- x}{- 1} = \frac{ln ( arccos ( ( D ^{2} - 3 D + 2 ) y ) + 2 πn )}{- 1}

Simplify

x = - ln (arccos (y (D^{2} - 3 D + 2)) + 2 πn)

x = - ln (arccos (y (D^{2} - 3 D + 2)) + 2 πn)

x = - ln (arccos (y (D^{2} - 3 D + 2)) + 2 πn)

Solve

e^{- x} = - arccos ((D^{2} - 3 D + 2) y) + 2 πn : x = - ln (- arccos (y (D^{2} - 3 D + 2)) + 2 πn)

e^{- x} = - arccos ((D^{2} - 3 D + 2) y) + 2 πn

Apply exponent rules

e^{- x} = - arccos ((D^{2} - 3 D + 2) y) + 2 πn

If

f (x) = g (x)

, then

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

ln (e^{- x}) = ln (- arccos ((D^{2} - 3 D + 2) y) + 2 πn)

Apply log rule:

ln (e^{a}) = a

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

- x = ln (- arccos ((D^{2} - 3 D + 2) y) + 2 πn)

- x = ln (- arccos ((D^{2} - 3 D + 2) y) + 2 πn)

Solve

- x = ln (- arccos ((D^{2} - 3 D + 2) y) + 2 πn) : x = - ln (- arccos (y (D^{2} - 3 D + 2)) + 2 πn)

- x = ln (- arccos ((D^{2} - 3 D + 2) y) + 2 πn)

Divide both sides by

- 1

- x = ln (- arccos ((D^{2} - 3 D + 2) y) + 2 πn)

Divide both sides by

- 1

\frac{- x}{- 1} = \frac{ln ( - arccos ( ( D ^{2} - 3 D + 2 ) y ) + 2 πn )}{- 1}

Simplify

x = - ln (- arccos (y (D^{2} - 3 D + 2)) + 2 πn)

x = - ln (- arccos (y (D^{2} - 3 D + 2)) + 2 πn)

x = - ln (- arccos (y (D^{2} - 3 D + 2)) + 2 πn)

x = - ln (arccos (y (D^{2} - 3 D + 2)) + 2 πn), x = - ln (- arccos (y (D^{2} - 3 D + 2)) + 2 πn)

Popular Examples

2+cos(2x)=-5sin(x)

2 + cos (2 x) = - 5 sin (x)

cos^2(θ)-1/4 =0

cos^{2} (θ) - \frac{1}{4} = 0

2cot(x)+sec^2(x)=0

2 cot (x) + sec^{2} (x) = 0

picos(θ)=-pi/2

π cos (θ) = - \frac{π}{2}

solvefor x,y=1+sin((pix)/4)solve for

x, y = 1 + sin (\frac{π x}{4})