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

The general solution for 3sin(θ)=1-sin(θ) is θ=0.25268…+2pin,θ=pi-0.25268…+2pin

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3sin(θ)=1-sin(θ)

3 sin (θ) = 1 - sin (θ)

θ = 0.25268 \dots + 2 πn, θ = π - 0.25268 \dots + 2 πn

Degrees

θ = 14.47751 \dots^{\circ} + 36 0^{\circ} n, θ = 165.52248 \dots^{\circ} + 36 0^{\circ} n

Solution steps

3 sin (θ) = 1 - sin (θ)

Solve by substitution

3 sin (θ) = 1 - sin (θ)

Let:

sin (θ) = u

3 u = 1 - u

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

3 u = 1 - u

Move

u

to the left side

3 u = 1 - u

Add

u

to both sides

3 u + u = 1 - u + u

Simplify

4 u = 1

4 u = 1

Divide both sides by

4

4 u = 1

Divide both sides by

4

\frac{4 u}{4} = \frac{1}{4}

Simplify

u = \frac{1}{4}

u = \frac{1}{4}

Substitute back

u = sin (θ)

sin (θ) = \frac{1}{4}

sin (θ) = \frac{1}{4}

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

sin (θ) = \frac{1}{4}

Apply trig inverse properties

sin (θ) = \frac{1}{4}

General solutions for

sin (θ) = \frac{1}{4}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

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

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

Combine all the solutions

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

Show solutions in decimal form

θ = 0.25268 \dots + 2 πn, θ = π - 0.25268 \dots + 2 πn

1 = sin (9 0^{\circ} - θ) - 0.075 cos (9 0^{\circ} - θ)

- 6 sin (θ) = - 32

cos (x) = \frac{4 ( \frac{1}{2} ) - 1}{3}

sin (x) = \frac{π}{4} + \frac{0 π}{2}

cos (2 x) - sin (x) - 1 = 0

What is the general solution for 3sin(θ)=1-sin(θ) ?
The general solution for 3sin(θ)=1-sin(θ) is θ=0.25268…+2pin,θ=pi-0.25268…+2pin