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

The general solution for 4cos(θ)=sqrt(2)+2cos(θ) is θ= pi/4+2pin,θ=(7pi)/4+2pin

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4cos(θ)=sqrt(2)+2cos(θ)

4 cos (θ) = 2 + 2 cos (θ)

θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

Degrees

θ = 4 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n

Solution steps

4 cos (θ) = 2 + 2 cos (θ)

Solve by substitution

4 cos (θ) = 2 + 2 cos (θ)

Let:

cos (θ) = u

4 u = 2 + 2 u

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

4 u = 2 + 2 u

Move

2 u

to the left side

4 u = 2 + 2 u

Subtract

2 u

from both sides

4 u - 2 u = 2 + 2 u - 2 u

Simplify

2 u = 2

2 u = 2

Divide both sides by

2

2 u = 2

Divide both sides by

2

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

Simplify

u = \frac{2}{2}

u = \frac{2}{2}

Substitute back

u = cos (θ)

cos (θ) = \frac{2}{2}

cos (θ) = \frac{2}{2}

cos (θ) = \frac{2}{2} : θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

cos (θ) = \frac{2}{2}

General solutions for

cos (θ) = \frac{2}{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}

θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

Combine all the solutions

θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

x, T (6) = 3.15 cos (\frac{π}{6} x) + 19.15

3 tan^{2} (x) = 1, - π \leq x \leq π

cos (2 t) - cos (t) = 0.5

0 = - 6 csc (x) cot (x)

0.08 = 0.12 cos^{2} (3.922 t)

What is the general solution for 4cos(θ)=sqrt(2)+2cos(θ) ?
The general solution for 4cos(θ)=sqrt(2)+2cos(θ) is θ= pi/4+2pin,θ=(7pi)/4+2pin