What is the general solution for 0=1-3cos(θ) ?

The general solution for 0=1-3cos(θ) is θ=1.23095…+2pin,θ=2pi-1.23095…+2pin

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0=1-3cos(θ)

0 = 1 - 3 cos (θ)

θ = 1.23095 \dots + 2 πn, θ = 2 π - 1.23095 \dots + 2 πn

Degrees

θ = 70.52877 \dots^{\circ} + 36 0^{\circ} n, θ = 289.47122 \dots^{\circ} + 36 0^{\circ} n

Solution steps

0 = 1 - 3 cos (θ)

Switch sides

1 - 3 cos (θ) = 0

Move

1

to the right side

1 - 3 cos (θ) = 0

Subtract

1

from both sides

1 - 3 cos (θ) - 1 = 0 - 1

Simplify

- 3 cos (θ) = - 1

- 3 cos (θ) = - 1

Divide both sides by

- 3

- 3 cos (θ) = - 1

Divide both sides by

- 3

\frac{- 3 cos ( θ )}{- 3} = \frac{- 1}{- 3}

Simplify

cos (θ) = \frac{1}{3}

cos (θ) = \frac{1}{3}

Apply trig inverse properties

cos (θ) = \frac{1}{3}

General solutions for

cos (θ) = \frac{1}{3}

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

θ = arccos (\frac{1}{3}) + 2 πn, θ = 2 π - arccos (\frac{1}{3}) + 2 πn

θ = arccos (\frac{1}{3}) + 2 πn, θ = 2 π - arccos (\frac{1}{3}) + 2 πn

Show solutions in decimal form

θ = 1.23095 \dots + 2 πn, θ = 2 π - 1.23095 \dots + 2 πn

(tan (\frac{3 a}{2})) tan (\frac{a}{2}) = 3

1 = 3 sin (x)

\frac{sin ( 11 5 ^{\circ} )}{20} = \frac{sin ( B )}{15}

3 tan (θ) + 1 = 2 tan (θ)

cos (x) = \frac{18}{25}

What is the general solution for 0=1-3cos(θ) ?
The general solution for 0=1-3cos(θ) is θ=1.23095…+2pin,θ=2pi-1.23095…+2pin