What is the general solution for 0=cos(x)(1-cos(x))+sin^2(x) ?

The general solution for 0=cos(x)(1-cos(x))+sin^2(x) is x=(2pi)/3+2pin,x=(4pi)/3+2pin,x=2pin

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

Solution

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

Solution

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = 2 πn

Degrees

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Switch sides

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

Rewrite using trig identities

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

Simplify

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

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

a = cos (x), b = 1, c = cos (x)

= cos (x) \cdot 1 - cos (x) cos (x)

= 1 \cdot cos (x) - cos (x) cos (x)

Simplify

1 \cdot cos (x) - cos (x) cos (x) : cos (x) - cos^{2} (x)

1 \cdot cos (x) - cos (x) cos (x)

1 \cdot cos (x) = cos (x)

1 \cdot cos (x)

Multiply:

1 \cdot cos (x) = cos (x)

= cos (x)

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

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

1 + u - 2 u^{2} = 0 : u = - \frac{1}{2}, u = 1

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

Write in the standard form

a x^{2} + b x + c = 0

- 2 u^{2} + u + 1 = 0

Solve with the quadratic formula

- 2 u^{2} + u + 1 = 0

Quadratic Equation Formula:

For

a = - 2, b = 1, c = 1

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 2 ) \cdot 1}{2 ( - 2 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 2 ) \cdot 1}{2 ( - 2 )}

1^{2} - 4 (- 2) \cdot 1 = 3

1^{2} - 4 (- 2) \cdot 1

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 2) \cdot 1

Apply rule

- (- a) = a

= 1 + 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- 1 \pm 3}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- 1 + 3}{2 ( - 2 )}, u_{2} = \frac{- 1 - 3}{2 ( - 2 )}

u = \frac{- 1 + 3}{2 ( - 2 )} : - \frac{1}{2}

\frac{- 1 + 3}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 3}{- 2 \cdot 2}

Add/Subtract the numbers:

- 1 + 3 = 2

= \frac{2}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{- 4}

Apply the fraction rule:

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

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

u = \frac{- 1 - 3}{2 ( - 2 )} : 1

\frac{- 1 - 3}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 3}{- 2 \cdot 2}

Subtract the numbers:

- 1 - 3 = - 4

= \frac{- 4}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{- 4}

Apply the fraction rule:

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

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = - \frac{1}{2}, u = 1

Substitute back

u = cos (x)

cos (x) = - \frac{1}{2}, cos (x) = 1

cos (x) = - \frac{1}{2}, cos (x) = 1

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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

General solutions for

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

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 = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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

Combine all the solutions

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 0=cos(x)(1-cos(x))+sin^2(x) ?
The general solution for 0=cos(x)(1-cos(x))+sin^2(x) is x=(2pi)/3+2pin,x=(4pi)/3+2pin,x=2pin