What is the general solution for 4cos(θ)=2cos(θ) ?

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

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4cos(θ)=2cos(θ)

4 cos (θ) = 2 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

4 cos (θ) = 2 cos (θ)

Solve by substitution

4 cos (θ) = 2 cos (θ)

Let:

cos (θ) = u

4 u = 2 u

4 u = 2 u : u = 0

4 u = 2 u

Move

2 u

to the left side

4 u = 2 u

Subtract

2 u

from both sides

4 u - 2 u = 2 u - 2 u

Simplify

2 u = 0

2 u = 0

Divide both sides by

2

2 u = 0

Divide both sides by

2

\frac{2 u}{2} = \frac{0}{2}

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

2 cos (θ) = - 2

x, cos (2 x) + \frac{2}{cos ( 2 x )} + 3 = 0

sin (t) = \frac{250}{25 0 ^{2} + 30 0 ^{2}}

0 = sin (13 x)

cos (a) = \frac{9}{41}

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