What is the general solution for 3cos(x)=cos(x)-1 ?

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

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3cos(x)=cos(x)-1

3 cos (x) = cos (x) - 1

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

3 cos (x) = cos (x) - 1

Solve by substitution

3 cos (x) = cos (x) - 1

Let:

cos (x) = u

3 u = u - 1

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

3 u = u - 1

Move

u

to the left side

3 u = u - 1

Subtract

u

from both sides

3 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

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

3 sin^{2} (x) = cos^{2} (x) - 4 cos (x)

cos (x) = 0.9224

- 9 sin^{2} (θ) + 3 cos (θ) - 2 = - 9

4 sin (x) = 4 cos (x)

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