What is the general solution for solvefor x,13y=cos^4(1-2x) ?

The general solution for solvefor x,13y=cos^4(1-2x) is

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Popular Trigonometry >

solvefor x,13y=cos^4(1-2x)

Solution

solvefor

Solution

x = - \frac{arccos ( 13 y )}{2} - πn + \frac{1}{2}, x = \frac{arccos ( 13 y )}{2} - πn + \frac{1}{2}, x = - \frac{arccos ( - 13 y )}{2} - πn + \frac{1}{2}, x = \frac{arccos ( - 13 y )}{2} - πn + \frac{1}{2}

Solution steps

13 y = cos^{4} (1 - 2 x)

Switch sides

cos^{4} (1 - 2 x) = 13 y

Solve by substitution

cos^{4} (1 - 2 x) = 13 y

Let:

cos (1 - 2 x) = u

u^{4} = 13 y

u^{4} = 13 y : u = 13 y, u = - 13 y, u = i 13 y, u = - i 13 y

u^{4} = 13 y

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

v^{2} = 13 y

Solve

v^{2} = 13 y : v = 13 y, v = - 13 y

v^{2} = 13 y

For

(g (x))^{2} = f (a)

the solutions are

g (x) = f (a), - f (a)

v = 13 y, v = - 13 y

v = 13 y, v = - 13 y

Substitute back

v = u^{2},

solve for

u

Solve

u^{2} = 13 y : u = 13 y, u = - 13 y

u^{2} = 13 y

Apply radical rule:

assuming

a \geq 0, b \geq 0

u^{2} = 13 y

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 13 y, u = - 13 y

Solve

u^{2} = - 13 y : u = i 13 y, u = - i 13 y

u^{2} = - 13 y

Apply radical rule:

assuming

a \geq 0, b \geq 0

u^{2} = - 13 y

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = - 13 y, u = - - 13 y

Simplify

- 13 y : i 13 y

- 13 y

Apply radical rule:

- a = - 1 a,

assuming

a \geq 0

- 13 y = - 1 13 y

= - 1 13 y

Apply imaginary number rule:

- 1 = i

= i 13 y

Simplify

- - 13 y : - i 13 y

- - 13 y

Simplify

- 13 y : i 13 y

- 13 y

Apply radical rule:

- a = - 1 a,

assuming

a \geq 0

- 13 y = - 1 13 y

= - 1 13 y

Apply imaginary number rule:

- 1 = i

= i 13 y

= - i 13 y

u = i 13 y, u = - i 13 y

The solutions are

u = 13 y, u = - 13 y, u = i 13 y, u = - i 13 y

Substitute back

u = cos (1 - 2 x)

cos (1 - 2 x) = 13 y, cos (1 - 2 x) = - 13 y, cos (1 - 2 x) = i 13 y, cos (1 - 2 x) = - i 13 y

cos (1 - 2 x) = 13 y, cos (1 - 2 x) = - 13 y, cos (1 - 2 x) = i 13 y, cos (1 - 2 x) = - i 13 y

cos (1 - 2 x) = 13 y : x = - \frac{arccos ( 13 y )}{2} - πn + \frac{1}{2}, x = \frac{arccos ( 13 y )}{2} - πn + \frac{1}{2}

cos (1 - 2 x) = 13 y

Apply trig inverse properties

cos (1 - 2 x) = 13 y

General solutions for

cos (1 - 2 x) = 13 y

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

1 - 2 x = arccos (13 y) + 2 πn, 1 - 2 x = - arccos (13 y) + 2 πn

1 - 2 x = arccos (13 y) + 2 πn, 1 - 2 x = - arccos (13 y) + 2 πn

Solve

1 - 2 x = arccos (13 y) + 2 πn : x = - \frac{arccos ( 13 y )}{2} - πn + \frac{1}{2}

1 - 2 x = arccos (13 y) + 2 πn

Move

1

to the right side

1 - 2 x = arccos (13 y) + 2 πn

Subtract

1

from both sides

1 - 2 x - 1 = arccos (13 y) + 2 πn - 1

Simplify

- 2 x = arccos (13 y) + 2 πn - 1

- 2 x = arccos (13 y) + 2 πn - 1

Divide both sides by

- 2

- 2 x = arccos (13 y) + 2 πn - 1

Divide both sides by

- 2

\frac{- 2 x}{- 2} = \frac{arccos ( 13 y )}{- 2} + \frac{2 πn}{- 2} - \frac{1}{- 2}

Simplify

\frac{- 2 x}{- 2} = \frac{arccos ( 13 y )}{- 2} + \frac{2 πn}{- 2} - \frac{1}{- 2}

Simplify

\frac{- 2 x}{- 2} : x

\frac{- 2 x}{- 2}

Apply the fraction rule:

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

= \frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{arccos ( 13 y )}{- 2} + \frac{2 πn}{- 2} - \frac{1}{- 2} : - \frac{arccos ( 13 y )}{2} - πn + \frac{1}{2}

\frac{arccos ( 13 y )}{- 2} + \frac{2 πn}{- 2} - \frac{1}{- 2}

Apply the fraction rule:

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

= - \frac{arccos ( 13 y )}{2} + \frac{2 πn}{- 2} - \frac{1}{- 2}

\frac{2 πn}{- 2} = - πn

\frac{2 πn}{- 2}

Apply the fraction rule:

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

= - \frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= - πn

= - \frac{arccos ( 13 y )}{2} - πn - \frac{1}{- 2}

Apply the fraction rule:

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

= - \frac{arccos ( 13 y )}{2} - πn - (- \frac{1}{2})

Apply rule

- (- a) = a

= - \frac{arccos ( 13 y )}{2} - πn + \frac{1}{2}

x = - \frac{arccos ( 13 y )}{2} - πn + \frac{1}{2}

x = - \frac{arccos ( 13 y )}{2} - πn + \frac{1}{2}

x = - \frac{arccos ( 13 y )}{2} - πn + \frac{1}{2}

Solve

1 - 2 x = - arccos (13 y) + 2 πn : x = \frac{arccos ( 13 y )}{2} - πn + \frac{1}{2}

1 - 2 x = - arccos (13 y) + 2 πn

Move

1

to the right side

1 - 2 x = - arccos (13 y) + 2 πn

Subtract

1

from both sides

1 - 2 x - 1 = - arccos (13 y) + 2 πn - 1

Simplify

- 2 x = - arccos (13 y) + 2 πn - 1

- 2 x = - arccos (13 y) + 2 πn - 1

Divide both sides by

- 2

- 2 x = - arccos (13 y) + 2 πn - 1

Divide both sides by

- 2

\frac{- 2 x}{- 2} = - \frac{arccos ( 13 y )}{- 2} + \frac{2 πn}{- 2} - \frac{1}{- 2}

Simplify

\frac{- 2 x}{- 2} = - \frac{arccos ( 13 y )}{- 2} + \frac{2 πn}{- 2} - \frac{1}{- 2}

Simplify

\frac{- 2 x}{- 2} : x

\frac{- 2 x}{- 2}

Apply the fraction rule:

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

= \frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

- \frac{arccos ( 13 y )}{- 2} + \frac{2 πn}{- 2} - \frac{1}{- 2} : \frac{arccos ( 13 y )}{2} - πn + \frac{1}{2}

- \frac{arccos ( 13 y )}{- 2} + \frac{2 πn}{- 2} - \frac{1}{- 2}

\frac{arccos ( 13 y )}{- 2} = - \frac{arccos ( 13 y )}{2}

\frac{arccos ( 13 y )}{- 2}

Apply the fraction rule:

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

= - \frac{arccos ( 13 y )}{2}

\frac{2 πn}{- 2} = - πn

\frac{2 πn}{- 2}

Apply the fraction rule:

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

= - \frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= - πn

= - - \frac{arccos ( 13 y )}{2} - πn - \frac{1}{- 2}

Apply rule

- (- a) = a

= \frac{arccos ( 13 y )}{2} - πn - \frac{1}{- 2}

Apply the fraction rule:

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

= \frac{arccos ( 13 y )}{2} - πn - (- \frac{1}{2})

Apply rule

- (- a) = a

= \frac{arccos ( 13 y )}{2} - πn + \frac{1}{2}

x = \frac{arccos ( 13 y )}{2} - πn + \frac{1}{2}

x = \frac{arccos ( 13 y )}{2} - πn + \frac{1}{2}

x = \frac{arccos ( 13 y )}{2} - πn + \frac{1}{2}

x = - \frac{arccos ( 13 y )}{2} - πn + \frac{1}{2}, x = \frac{arccos ( 13 y )}{2} - πn + \frac{1}{2}

cos (1 - 2 x) = - 13 y : x = - \frac{arccos ( - 13 y )}{2} - πn + \frac{1}{2}, x = \frac{arccos ( - 13 y )}{2} - πn + \frac{1}{2}

cos (1 - 2 x) = - 13 y

Apply trig inverse properties

cos (1 - 2 x) = - 13 y

General solutions for

cos (1 - 2 x) = - 13 y

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

1 - 2 x = arccos (- 13 y) + 2 πn, 1 - 2 x = - arccos (- 13 y) + 2 πn

1 - 2 x = arccos (- 13 y) + 2 πn, 1 - 2 x = - arccos (- 13 y) + 2 πn

Solve

1 - 2 x = arccos (- 13 y) + 2 πn : x = - \frac{arccos ( - 13 y )}{2} - πn + \frac{1}{2}

1 - 2 x = arccos (- 13 y) + 2 πn

Move

1

to the right side

1 - 2 x = arccos (- 13 y) + 2 πn

Subtract

1

from both sides

1 - 2 x - 1 = arccos (- 13 y) + 2 πn - 1

Simplify

- 2 x = arccos (- 13 y) + 2 πn - 1

- 2 x = arccos (- 13 y) + 2 πn - 1

Divide both sides by

- 2

- 2 x = arccos (- 13 y) + 2 πn - 1

Divide both sides by

- 2

\frac{- 2 x}{- 2} = \frac{arccos ( - 13 y )}{- 2} + \frac{2 πn}{- 2} - \frac{1}{- 2}

Simplify

\frac{- 2 x}{- 2} = \frac{arccos ( - 13 y )}{- 2} + \frac{2 πn}{- 2} - \frac{1}{- 2}

Simplify

\frac{- 2 x}{- 2} : x

\frac{- 2 x}{- 2}

Apply the fraction rule:

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

= \frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{arccos ( - 13 y )}{- 2} + \frac{2 πn}{- 2} - \frac{1}{- 2} : - \frac{arccos ( - 13 y )}{2} - πn + \frac{1}{2}

\frac{arccos ( - 13 y )}{- 2} + \frac{2 πn}{- 2} - \frac{1}{- 2}

Apply the fraction rule:

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

= - \frac{arccos ( - 13 y )}{2} + \frac{2 πn}{- 2} - \frac{1}{- 2}

\frac{2 πn}{- 2} = - πn

\frac{2 πn}{- 2}

Apply the fraction rule:

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

= - \frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= - πn

= - \frac{arccos ( - 13 y )}{2} - πn - \frac{1}{- 2}

Apply the fraction rule:

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

= - \frac{arccos ( - 13 y )}{2} - πn - (- \frac{1}{2})

Apply rule

- (- a) = a

= - \frac{arccos ( - 13 y )}{2} - πn + \frac{1}{2}

x = - \frac{arccos ( - 13 y )}{2} - πn + \frac{1}{2}

x = - \frac{arccos ( - 13 y )}{2} - πn + \frac{1}{2}

x = - \frac{arccos ( - 13 y )}{2} - πn + \frac{1}{2}

Solve

1 - 2 x = - arccos (- 13 y) + 2 πn : x = \frac{arccos ( - 13 y )}{2} - πn + \frac{1}{2}

1 - 2 x = - arccos (- 13 y) + 2 πn

Move

1

to the right side

1 - 2 x = - arccos (- 13 y) + 2 πn

Subtract

1

from both sides

1 - 2 x - 1 = - arccos (- 13 y) + 2 πn - 1

Simplify

- 2 x = - arccos (- 13 y) + 2 πn - 1

- 2 x = - arccos (- 13 y) + 2 πn - 1

Divide both sides by

- 2

- 2 x = - arccos (- 13 y) + 2 πn - 1

Divide both sides by

- 2

\frac{- 2 x}{- 2} = - \frac{arccos ( - 13 y )}{- 2} + \frac{2 πn}{- 2} - \frac{1}{- 2}

Simplify

\frac{- 2 x}{- 2} = - \frac{arccos ( - 13 y )}{- 2} + \frac{2 πn}{- 2} - \frac{1}{- 2}

Simplify

\frac{- 2 x}{- 2} : x

\frac{- 2 x}{- 2}

Apply the fraction rule:

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

= \frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

- \frac{arccos ( - 13 y )}{- 2} + \frac{2 πn}{- 2} - \frac{1}{- 2} : \frac{arccos ( - 13 y )}{2} - πn + \frac{1}{2}

- \frac{arccos ( - 13 y )}{- 2} + \frac{2 πn}{- 2} - \frac{1}{- 2}

\frac{arccos ( - 13 y )}{- 2} = - \frac{arccos ( - 13 y )}{2}

\frac{arccos ( - 13 y )}{- 2}

Apply the fraction rule:

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

= - \frac{arccos ( - 13 y )}{2}

\frac{2 πn}{- 2} = - πn

\frac{2 πn}{- 2}

Apply the fraction rule:

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

= - \frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= - πn

= - - \frac{arccos ( - 13 y )}{2} - πn - \frac{1}{- 2}

Apply rule

- (- a) = a

= \frac{arccos ( - 13 y )}{2} - πn - \frac{1}{- 2}

Apply the fraction rule:

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

= \frac{arccos ( - 13 y )}{2} - πn - (- \frac{1}{2})

Apply rule

- (- a) = a

= \frac{arccos ( - 13 y )}{2} - πn + \frac{1}{2}

x = \frac{arccos ( - 13 y )}{2} - πn + \frac{1}{2}

x = \frac{arccos ( - 13 y )}{2} - πn + \frac{1}{2}

x = \frac{arccos ( - 13 y )}{2} - πn + \frac{1}{2}

x = - \frac{arccos ( - 13 y )}{2} - πn + \frac{1}{2}, x = \frac{arccos ( - 13 y )}{2} - πn + \frac{1}{2}

cos (1 - 2 x) = i 13 y :

No Solution

cos (1 - 2 x) = i 13 y

No Solution

cos (1 - 2 x) = - i 13 y :

No Solution

cos (1 - 2 x) = - i 13 y

No Solution

Combine all the solutions

x = - \frac{arccos ( 13 y )}{2} - πn + \frac{1}{2}, x = \frac{arccos ( 13 y )}{2} - πn + \frac{1}{2}, x = - \frac{arccos ( - 13 y )}{2} - πn + \frac{1}{2}, x = \frac{arccos ( - 13 y )}{2} - πn + \frac{1}{2}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for solvefor x,13y=cos^4(1-2x) ?
The general solution for solvefor x,13y=cos^4(1-2x) is