What is the general solution for (sin^3(x))/(cos^3(x)-cos(x))=tan(x) ?

The general solution for (sin^3(x))/(cos^3(x)-cos(x))=tan(x) is No Solution for x\in\mathbb{R}

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(sin^3(x))/(cos^3(x)-cos(x))=tan(x)

Solution

\frac{sin ^{3} ( x )}{cos ^{3} ( x ) - cos ( x )} = tan (x)

Solution

No Solution for x \in R

Solution steps

\frac{sin ^{3} ( x )}{cos ^{3} ( x ) - cos ( x )} = tan (x)

Subtract

tan (x)

from both sides

\frac{sin ^{3} ( x )}{cos ^{3} ( x ) - cos ( x )} - tan (x) = 0

Simplify

\frac{sin ^{3} ( x )}{cos ^{3} ( x ) - cos ( x )} - tan (x) : \frac{sin ^{3} ( x ) - tan ( x ) ( cos ^{3} ( x ) - cos ( x ) )}{cos ^{3} ( x ) - cos ( x )}

\frac{sin ^{3} ( x )}{cos ^{3} ( x ) - cos ( x )} - tan (x)

Convert element to fraction:

tan (x) = \frac{tan ( x ) ( cos ^{3} ( x ) - cos ( x ) )}{cos ^{3} ( x ) - cos ( x )}

= \frac{sin ^{3} ( x )}{cos ^{3} ( x ) - cos ( x )} - \frac{tan ( x ) ( cos ^{3} ( x ) - cos ( x ) )}{cos ^{3} ( x ) - cos ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{sin ^{3} ( x ) - tan ( x ) ( cos ^{3} ( x ) - cos ( x ) )}{cos ^{3} ( x ) - cos ( x )}

\frac{sin ^{3} ( x ) - tan ( x ) ( cos ^{3} ( x ) - cos ( x ) )}{cos ^{3} ( x ) - cos ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sin^{3} (x) - tan (x) (cos^{3} (x) - cos (x)) = 0

Rewrite using trig identities

sin^{3} (x) - (- cos (x) + cos^{3} (x)) tan (x)

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= sin^{3} (x) - (- cos (x) + cos^{3} (x)) \frac{sin ( x )}{cos ( x )}

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

(- cos (x) + cos^{3} (x)) \frac{sin ( x )}{cos ( x )}

Multiply fractions:

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

= \frac{sin ( x ) ( - cos ( x ) + cos ^{3} ( x ) )}{cos ( x )}

Factor

- cos (x) + cos^{3} (x) : cos (x) (- 1 + cos^{2} (x))

- cos (x) + cos^{3} (x)

Apply exponent rule:

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

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

= - cos (x) + cos (x) cos^{2} (x)

Factor out common term

cos (x)

= cos (x) (- 1 + cos^{2} (x))

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

Cancel the common factor:

cos (x)

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

= sin^{3} (x) - sin (x) (cos^{2} (x) - 1)

sin^{3} (x) - (- 1 + cos^{2} (x)) sin (x) = 0

Factor

sin^{3} (x) - (- 1 + cos^{2} (x)) sin (x) : sin (x) (sin^{2} (x) - (cos (x) + 1) (cos (x) - 1))

sin^{3} (x) - (- 1 + cos^{2} (x)) sin (x)

Apply exponent rule:

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

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

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

Factor out common term

sin (x)

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

Factor

sin^{2} (x) - (cos^{2} (x) - 1) : sin^{2} (x) - (cos (x) + 1) (cos (x) - 1)

sin^{2} (x) - (cos^{2} (x) - 1)

Factor

cos^{2} (x) - 1 : (cos (x) + 1) (cos (x) - 1)

cos^{2} (x) - 1

Rewrite

1

1^{2}

= cos^{2} (x) - 1^{2}

Apply Difference of Two Squares Formula:

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

cos^{2} (x) - 1^{2} = (cos (x) + 1) (cos (x) - 1)

= (cos (x) + 1) (cos (x) - 1)

= sin^{2} (x) - (cos (x) + 1) (cos (x) - 1)

= sin (x) (sin^{2} (x) - (cos (x) + 1) (cos (x) - 1))

sin (x) (sin^{2} (x) - (cos (x) + 1) (cos (x) - 1)) = 0

Solving each part separately

sin (x) = 0 or sin^{2} (x) - (cos (x) + 1) (cos (x) - 1) = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

sin^{2} (x) - (cos (x) + 1) (cos (x) - 1) = 0 : x = 2 πn, x = π + 2 πn

sin^{2} (x) - (cos (x) + 1) (cos (x) - 1) = 0

Rewrite using trig identities

sin^{2} (x) - (- 1 + cos (x)) (1 + cos (x))

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

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

= 1 - cos^{2} (x) - (- 1 + cos (x)) (1 + cos (x))

Simplify

1 - cos^{2} (x) - (- 1 + cos (x)) (1 + cos (x)) : - 2 cos^{2} (x) + 2

1 - cos^{2} (x) - (- 1 + cos (x)) (1 + cos (x))

Expand

- (- 1 + cos (x)) (1 + cos (x)) : - cos^{2} (x) + 1

Expand

(- 1 + cos (x)) (1 + cos (x)) : cos^{2} (x) - 1

(- 1 + cos (x)) (1 + cos (x))

Apply Difference of Two Squares Formula:

(a - b) (a + b) = a^{2} - b^{2}

a = cos (x), b = 1

= cos^{2} (x) - 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= cos^{2} (x) - 1

= - (cos^{2} (x) - 1)

Distribute parentheses

= - (cos^{2} (x)) - (- 1)

Apply minus-plus rules

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

= - cos^{2} (x) + 1

= 1 - cos^{2} (x) - cos^{2} (x) + 1

Simplify

1 - cos^{2} (x) - cos^{2} (x) + 1 : - 2 cos^{2} (x) + 2

1 - cos^{2} (x) - cos^{2} (x) + 1

Add similar elements:

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

= 1 - 2 cos^{2} (x) + 1

Group like terms

= - 2 cos^{2} (x) + 1 + 1

Add the numbers:

1 + 1 = 2

= - 2 cos^{2} (x) + 2

= - 2 cos^{2} (x) + 2

= - 2 cos^{2} (x) + 2

2 - 2 cos^{2} (x) = 0

Solve by substitution

2 - 2 cos^{2} (x) = 0

Let:

cos (x) = u

2 - 2 u^{2} = 0

2 - 2 u^{2} = 0 : u = 1, u = - 1

2 - 2 u^{2} = 0

Move

2

to the right side

2 - 2 u^{2} = 0

Subtract

2

from both sides

2 - 2 u^{2} - 2 = 0 - 2

Simplify

- 2 u^{2} = - 2

- 2 u^{2} = - 2

Divide both sides by

- 2

- 2 u^{2} = - 2

Divide both sides by

- 2

\frac{- 2 u ^{2}}{- 2} = \frac{- 2}{- 2}

Simplify

u^{2} = 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

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

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

Substitute back

u = cos (x)

cos (x) = 1, cos (x) = - 1

cos (x) = 1, cos (x) = - 1

cos (x) = 1 : x = 2 πn

cos (x) = 1

General solutions for

cos (x) = 1

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

General solutions for

cos (x) = - 1

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = π + 2 πn

x = π + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn

Since the equation is undefined for:

2 πn, π + 2 πn

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for (sin^3(x))/(cos^3(x)-cos(x))=tan(x) ?
The general solution for (sin^3(x))/(cos^3(x)-cos(x))=tan(x) is No Solution for x\in\mathbb{R}