What is the general solution for 4cos(x)=4-4cos(x) ?

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

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4cos(x)=4-4cos(x)

4 cos (x) = 4 - 4 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) = 4 - 4 cos (x)

Solve by substitution

4 cos (x) = 4 - 4 cos (x)

Let:

cos (x) = u

4 u = 4 - 4 u

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

4 u = 4 - 4 u

Move

4 u

to the left side

4 u = 4 - 4 u

Add

4 u

to both sides

4 u + 4 u = 4 - 4 u + 4 u

Simplify

8 u = 4

8 u = 4

Divide both sides by

8

8 u = 4

Divide both sides by

8

\frac{8 u}{8} = \frac{4}{8}

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

cos (2 x) - cos (6 x) - sin (4 x) = 0, 0 \leq x \leq π

0 = 3 sin (θ)

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

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

cos^{2} (x) - 4 cos (x) - 5 = 0

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