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

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

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1-cos(x)=2+cos(x)

1 - cos (x) = 2 + cos (x)

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

Degrees

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

Solution steps

1 - cos (x) = 2 + cos (x)

Solve by substitution

1 - cos (x) = 2 + cos (x)

Let:

cos (x) = u

1 - u = 2 + u

1 - u = 2 + u : u = - \frac{1}{2}

1 - u = 2 + u

Move

1

to the right side

1 - u = 2 + u

Subtract

1

from both sides

1 - u - 1 = 2 + u - 1

Simplify

- u = u + 1

- u = u + 1

Move

u

to the left side

- u = u + 1

Subtract

u

from both sides

- u - u = u + 1 - u

Simplify

- 2 u = 1

- 2 u = 1

Divide both sides by

- 2

- 2 u = 1

Divide both sides by

- 2

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

Simplify

u = - \frac{1}{2}

u = - \frac{1}{2}

Substitute back

u = cos (x)

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

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

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

Combine all the solutions

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

(1 + cos (4 x)) sin (4 x) = 2 cos^{2} (2 x)

\frac{5 sin ( x ) + 6}{sin ( x )} = 17

tan^{2} (x) sec (x) - 1 = 0

sin^{2} (x) + cos^{4} (x) = 2

(1 + sin^{2} (x)) = (4 cos (x))^{2}

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