What is the general solution for 8cos(θ)=3-4cos(θ) ?

The general solution for 8cos(θ)=3-4cos(θ) is θ=1.31811…+2pin,θ=2pi-1.31811…+2pin

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8cos(θ)=3-4cos(θ)

8 cos (θ) = 3 - 4 cos (θ)

θ = 1.31811 \dots + 2 πn, θ = 2 π - 1.31811 \dots + 2 πn

Degrees

θ = 75.52248 \dots^{\circ} + 36 0^{\circ} n, θ = 284.47751 \dots^{\circ} + 36 0^{\circ} n

Solution steps

8 cos (θ) = 3 - 4 cos (θ)

Solve by substitution

8 cos (θ) = 3 - 4 cos (θ)

Let:

cos (θ) = u

8 u = 3 - 4 u

8 u = 3 - 4 u : u = \frac{1}{4}

8 u = 3 - 4 u

Move

4 u

to the left side

8 u = 3 - 4 u

Add

4 u

to both sides

8 u + 4 u = 3 - 4 u + 4 u

Simplify

12 u = 3

12 u = 3

Divide both sides by

12

12 u = 3

Divide both sides by

12

\frac{12 u}{12} = \frac{3}{12}

Simplify

u = \frac{1}{4}

u = \frac{1}{4}

Substitute back

u = cos (θ)

cos (θ) = \frac{1}{4}

cos (θ) = \frac{1}{4}

cos (θ) = \frac{1}{4} : θ = arccos (\frac{1}{4}) + 2 πn, θ = 2 π - arccos (\frac{1}{4}) + 2 πn

cos (θ) = \frac{1}{4}

Apply trig inverse properties

cos (θ) = \frac{1}{4}

General solutions for

cos (θ) = \frac{1}{4}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

θ = arccos (\frac{1}{4}) + 2 πn, θ = 2 π - arccos (\frac{1}{4}) + 2 πn

θ = arccos (\frac{1}{4}) + 2 πn, θ = 2 π - arccos (\frac{1}{4}) + 2 πn

Combine all the solutions

θ = arccos (\frac{1}{4}) + 2 πn, θ = 2 π - arccos (\frac{1}{4}) + 2 πn

Show solutions in decimal form

θ = 1.31811 \dots + 2 πn, θ = 2 π - 1.31811 \dots + 2 πn

cos (θ) = \frac{3}{2}, sin (θ)

sin (θ) = 0.788

arctan (θ) = - 3

cos (θ) = \frac{2}{6}

0 = cos^{2} (x) + cos (x) - 1

What is the general solution for 8cos(θ)=3-4cos(θ) ?
The general solution for 8cos(θ)=3-4cos(θ) is θ=1.31811…+2pin,θ=2pi-1.31811…+2pin