What is the general solution for -cos(θ)+1=2cos(θ)+3 ?

The general solution for -cos(θ)+1=2cos(θ)+3 is θ=2.30052…+2pin,θ=-2.30052…+2pin

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-cos(θ)+1=2cos(θ)+3

- cos (θ) + 1 = 2 cos (θ) + 3

θ = 2.30052 \dots + 2 πn, θ = - 2.30052 \dots + 2 πn

Degrees

θ = 131.81031 \dots^{\circ} + 36 0^{\circ} n, θ = - 131.81031 \dots^{\circ} + 36 0^{\circ} n

Solution steps

- cos (θ) + 1 = 2 cos (θ) + 3

Solve by substitution

- cos (θ) + 1 = 2 cos (θ) + 3

Let:

cos (θ) = u

- u + 1 = 2 u + 3

- u + 1 = 2 u + 3 : u = - \frac{2}{3}

- u + 1 = 2 u + 3

Move

1

to the right side

- u + 1 = 2 u + 3

Subtract

1

from both sides

- u + 1 - 1 = 2 u + 3 - 1

Simplify

- u = 2 u + 2

- u = 2 u + 2

Move

2 u

to the left side

- u = 2 u + 2

Subtract

2 u

from both sides

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

Simplify

- 3 u = 2

- 3 u = 2

Divide both sides by

- 3

- 3 u = 2

Divide both sides by

- 3

\frac{- 3 u}{- 3} = \frac{2}{- 3}

Simplify

u = - \frac{2}{3}

u = - \frac{2}{3}

Substitute back

u = cos (θ)

cos (θ) = - \frac{2}{3}

cos (θ) = - \frac{2}{3}

cos (θ) = - \frac{2}{3} : θ = arccos (- \frac{2}{3}) + 2 πn, θ = - arccos (- \frac{2}{3}) + 2 πn

cos (θ) = - \frac{2}{3}

Apply trig inverse properties

cos (θ) = - \frac{2}{3}

General solutions for

cos (θ) = - \frac{2}{3}

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

θ = arccos (- \frac{2}{3}) + 2 πn, θ = - arccos (- \frac{2}{3}) + 2 πn

θ = arccos (- \frac{2}{3}) + 2 πn, θ = - arccos (- \frac{2}{3}) + 2 πn

Combine all the solutions

θ = arccos (- \frac{2}{3}) + 2 πn, θ = - arccos (- \frac{2}{3}) + 2 πn

Show solutions in decimal form

θ = 2.30052 \dots + 2 πn, θ = - 2.30052 \dots + 2 πn

cot^{2} (x) + cot (x) = 0

3 cos (2 θ) + 9 cos (θ) - 4 = - 2 cos (θ)

16 cos^{2} (x) - 12 = 0

cos (2 θ) - cos^{2} (θ) = 0

3 tan^{2} (x) + 5 tan (x) - 4 = 0

What is the general solution for -cos(θ)+1=2cos(θ)+3 ?
The general solution for -cos(θ)+1=2cos(θ)+3 is θ=2.30052…+2pin,θ=-2.30052…+2pin