What is the general solution for 2cos(θ)=cos(θ) ?

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

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2cos(θ)=cos(θ)

2 cos (θ) = cos (θ)

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

2 cos (θ) = cos (θ)

Solve by substitution

2 cos (θ) = cos (θ)

Let:

cos (θ) = u

2 u = u

2 u = u : u = 0

2 u = u

Move

u

to the left side

2 u = u

Subtract

u

from both sides

2 u - u = u - u

Simplify

u = 0

u = 0

Substitute back

u = cos (θ)

cos (θ) = 0

cos (θ) = 0

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = 0

General solutions for

cos (θ) = 0

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}

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

sin^{2} (x) \cdot cos (x) + 1 = 0

sin (B) = \frac{1}{2}, a = 170

sin (x - \frac{5 π}{3}) + sin (x + \frac{5 π}{3}) = 1

- sec (\frac{x}{2}) = 2 csc (\frac{x}{2})

sec^{2} (θ) - sec (θ) = 2, θ [0, \frac{π}{2}]

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