What is the general solution for 2cos(θ)-3=5cos(θ)-5 ?

The general solution for 2cos(θ)-3=5cos(θ)-5 is θ=0.84106…+2pin,θ=2pi-0.84106…+2pin

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2cos(θ)-3=5cos(θ)-5

2 cos (θ) - 3 = 5 cos (θ) - 5

θ = 0.84106 \dots + 2 πn, θ = 2 π - 0.84106 \dots + 2 πn

Degrees

θ = 48.18968 \dots^{\circ} + 36 0^{\circ} n, θ = 311.81031 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 cos (θ) - 3 = 5 cos (θ) - 5

Solve by substitution

2 cos (θ) - 3 = 5 cos (θ) - 5

Let:

cos (θ) = u

2 u - 3 = 5 u - 5

2 u - 3 = 5 u - 5 : u = \frac{2}{3}

2 u - 3 = 5 u - 5

Move

3

to the right side

2 u - 3 = 5 u - 5

Add

3

to both sides

2 u - 3 + 3 = 5 u - 5 + 3

Simplify

2 u = 5 u - 2

2 u = 5 u - 2

Move

5 u

to the left side

2 u = 5 u - 2

Subtract

5 u

from both sides

2 u - 5 u = 5 u - 2 - 5 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 = 2 π - 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 = 2 π - arccos (a) + 2 πn

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

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

Combine all the solutions

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

Show solutions in decimal form

θ = 0.84106 \dots + 2 πn, θ = 2 π - 0.84106 \dots + 2 πn

tan (θ) = \frac{4}{5}

tan^{2} (x) - sec (x) - 1 = 0

7 sin (x) cos (x) + 9 cos (x) = 0

2 cos (3 x) + 1 = 0

4 cos^{2} (x) - 3 = 0, 0 \leq x \leq 2 π

What is the general solution for 2cos(θ)-3=5cos(θ)-5 ?
The general solution for 2cos(θ)-3=5cos(θ)-5 is θ=0.84106…+2pin,θ=2pi-0.84106…+2pin