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

The general solution for d^2(1+cos(x))-(1+cos(x))^2=sin^2(x) is x=pi+2pin

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

Solution

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

Solution

x = π + 2 πn

Degrees

x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

sin^{2} (x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Group like terms

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

Subtract the numbers:

- 1 - 1 = - 2

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

- 2 + d^{2} - 2 u + u d^{2} = 0 : u = - 1; d \neq = 2, d \neq = - 2

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

Move

d^{2}

to the right side

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

Subtract

d^{2}

from both sides

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

Simplify

- 2 - 2 u + u d^{2} = - d^{2}

- 2 - 2 u + u d^{2} = - d^{2}

Move

2

to the right side

- 2 - 2 u + u d^{2} = - d^{2}

Add

2

to both sides

- 2 - 2 u + u d^{2} + 2 = - d^{2} + 2

Simplify

- 2 u + u d^{2} = - d^{2} + 2

- 2 u + u d^{2} = - d^{2} + 2

Factor

- 2 u + u d^{2} : u (- 2 + d^{2})

- 2 u + u d^{2}

Factor out common term

u

= u (- 2 + d^{2})

u (- 2 + d^{2}) = - d^{2} + 2

Divide both sides by

- 2 + d^{2}; d \neq = 2, d \neq = - 2

u (- 2 + d^{2}) = - d^{2} + 2

Divide both sides by

- 2 + d^{2}; d \neq = 2, d \neq = - 2

\frac{u ( - 2 + d ^{2} )}{- 2 + d ^{2}} = - \frac{d ^{2}}{- 2 + d ^{2}} + \frac{2}{- 2 + d ^{2}}; d \neq = 2, d \neq = - 2

Simplify

\frac{u ( - 2 + d ^{2} )}{- 2 + d ^{2}} = - \frac{d ^{2}}{- 2 + d ^{2}} + \frac{2}{- 2 + d ^{2}}

Simplify

\frac{u ( - 2 + d ^{2} )}{- 2 + d ^{2}} : u

\frac{u ( - 2 + d ^{2} )}{- 2 + d ^{2}}

Cancel the common factor:

- 2 + d^{2}

= u

Simplify

- \frac{d ^{2}}{- 2 + d ^{2}} + \frac{2}{- 2 + d ^{2}} : - 1

- \frac{d ^{2}}{- 2 + d ^{2}} + \frac{2}{- 2 + d ^{2}}

Apply rule

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

= \frac{- d ^{2} + 2}{- 2 + d ^{2}}

Factor out

- 1

= \frac{- ( d ^{2} - 2 )}{d ^{2} - 2}

Refine

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

Apply rule

\frac{a}{a} = 1

= - 1

u = - 1; d \neq = 2, d \neq = - 2

u = - 1; d \neq = 2, d \neq = - 2

u = - 1; d \neq = 2, d \neq = - 2

Substitute back

u = cos (x)

cos (x) = - 1; d \neq = 2, d \neq = - 2

cos (x) = - 1; d \neq = 2, d \neq = - 2

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 = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for d^2(1+cos(x))-(1+cos(x))^2=sin^2(x) ?
The general solution for d^2(1+cos(x))-(1+cos(x))^2=sin^2(x) is x=pi+2pin