What is the general solution for 4cos(x)=1+2cos(x) ?

The general solution for 4cos(x)=1+2cos(x) is x= pi/3+2pin,x=(5pi)/3+2pin

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4cos(x)=1+2cos(x)

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

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

4 u = 1 + 2 u

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

4 u = 1 + 2 u

Move

2 u

to the left side

4 u = 1 + 2 u

Subtract

2 u

from both sides

4 u - 2 u = 1 + 2 u - 2 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{π}{3} + 2 πn, x = \frac{5 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{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Combine all the solutions

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

2 sin (x) + 1 = csc (x)

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

7 sin (x) + 3 = sin (x)

2 cos (2 x) = 3

cos (2 x) = 1 + 4 sin (x)

What is the general solution for 4cos(x)=1+2cos(x) ?
The general solution for 4cos(x)=1+2cos(x) is x= pi/3+2pin,x=(5pi)/3+2pin