What is the general solution for -2cos(θ)+1=5cos(θ)+0 ?

The general solution for -2cos(θ)+1=5cos(θ)+0 is θ=1.42744…+2pin,θ=2pi-1.42744…+2pin

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-2cos(θ)+1=5cos(θ)+0

- 2 cos (θ) + 1 = 5 cos (θ) + 0

θ = 1.42744 \dots + 2 πn, θ = 2 π - 1.42744 \dots + 2 πn

Degrees

θ = 81.78678 \dots^{\circ} + 36 0^{\circ} n, θ = 278.21321 \dots^{\circ} + 36 0^{\circ} n

Solution steps

- 2 cos (θ) + 1 = 5 cos (θ) + 0

Solve by substitution

- 2 cos (θ) + 1 = 5 cos (θ) + 0

Let:

cos (θ) = u

- 2 u + 1 = 5 u + 0

- 2 u + 1 = 5 u + 0 : u = \frac{1}{7}

- 2 u + 1 = 5 u + 0

5 u + 0 = 5 u

- 2 u + 1 = 5 u

Move

1

to the right side

- 2 u + 1 = 5 u

Subtract

1

from both sides

- 2 u + 1 - 1 = 5 u - 1

Simplify

- 2 u = 5 u - 1

- 2 u = 5 u - 1

Move

5 u

to the left side

- 2 u = 5 u - 1

Subtract

5 u

from both sides

- 2 u - 5 u = 5 u - 1 - 5 u

Simplify

- 7 u = - 1

- 7 u = - 1

Divide both sides by

- 7

- 7 u = - 1

Divide both sides by

- 7

\frac{- 7 u}{- 7} = \frac{- 1}{- 7}

Simplify

u = \frac{1}{7}

u = \frac{1}{7}

Substitute back

u = cos (θ)

cos (θ) = \frac{1}{7}

cos (θ) = \frac{1}{7}

cos (θ) = \frac{1}{7} : θ = arccos (\frac{1}{7}) + 2 πn, θ = 2 π - arccos (\frac{1}{7}) + 2 πn

cos (θ) = \frac{1}{7}

Apply trig inverse properties

cos (θ) = \frac{1}{7}

General solutions for

cos (θ) = \frac{1}{7}

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

θ = arccos (\frac{1}{7}) + 2 πn, θ = 2 π - arccos (\frac{1}{7}) + 2 πn

θ = arccos (\frac{1}{7}) + 2 πn, θ = 2 π - arccos (\frac{1}{7}) + 2 πn

Combine all the solutions

θ = arccos (\frac{1}{7}) + 2 πn, θ = 2 π - arccos (\frac{1}{7}) + 2 πn

Show solutions in decimal form

θ = 1.42744 \dots + 2 πn, θ = 2 π - 1.42744 \dots + 2 πn

2 sin^{2} (θ) - sin (θ) = 0, 0 \leq θ < 2 π

cos (x^{2}) = \frac{1}{2}

sec^{2} (x) - 7 sec (x) = 0

e^{\frac{π}{2}} - e^{a r c t a n (1 + x^{2})} = 0

3 cot (θ) - 7 = 0

What is the general solution for -2cos(θ)+1=5cos(θ)+0 ?
The general solution for -2cos(θ)+1=5cos(θ)+0 is θ=1.42744…+2pin,θ=2pi-1.42744…+2pin