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

The general solution for 2cos(x)+1=cos(x) is x=pi+2pin

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

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

x = π + 2 πn

Degrees

x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

2 u + 1 = u

2 u + 1 = u : u = - 1

2 u + 1 = u

Move

1

to the right side

2 u + 1 = u

Subtract

1

from both sides

2 u + 1 - 1 = u - 1

Simplify

2 u = u - 1

2 u = u - 1

Move

u

to the left side

2 u = u - 1

Subtract

u

from both sides

2 u - u = u - 1 - u

Simplify

u = - 1

u = - 1

Substitute back

u = cos (x)

cos (x) = - 1

cos (x) = - 1

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

General solutions for

cos (x) = - 1

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 = π + 2 πn

x = π + 2 πn

Combine all the solutions

x = π + 2 πn

sin (\frac{( 60 )}{7}) = sin (\frac{( 81 )}{a})

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

sin (\frac{x}{3}) = - 0.85

2 cos (3 θ) = 2

tan (θ) = 2, 0 < θ < (\frac{π}{2}), sin (\frac{θ}{2})

What is the general solution for 2cos(x)+1=cos(x) ?
The general solution for 2cos(x)+1=cos(x) is x=pi+2pin