What is the general solution for 3cos(x)=2-cos(x) ?

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

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3cos(x)=2-cos(x)

3 cos (x) = 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

3 cos (x) = 2 - cos (x)

Solve by substitution

3 cos (x) = 2 - cos (x)

Let:

cos (x) = u

3 u = 2 - u

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

3 u = 2 - u

Move

u

to the left side

3 u = 2 - u

Add

u

to both sides

3 u + u = 2 - u + u

Simplify

4 u = 2

4 u = 2

Divide both sides by

4

4 u = 2

Divide both sides by

4

\frac{4 u}{4} = \frac{2}{4}

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 cos^{2} (3 x) - 3 cos (3 x) = - 1, 0 \leq x \leq 2 π

x, z = y^{s i n (x)}

t, 10 = 14 + 8 sin (\frac{π t}{12})

cos (4 x) cos (x - 1) = 0

cos (0.5 t + 0.464) = 1

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