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

The general solution for 1+sin(θ)=3sin(θ) is θ= pi/6+2pin,θ=(5pi)/6+2pin

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1+sin(θ)=3sin(θ)

1 + sin (θ) = 3 sin (θ)

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Degrees

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

1 + sin (θ) = 3 sin (θ)

Solve by substitution

1 + sin (θ) = 3 sin (θ)

Let:

sin (θ) = u

1 + u = 3 u

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

1 + u = 3 u

Move

1

to the right side

1 + u = 3 u

Subtract

1

from both sides

1 + u - 1 = 3 u - 1

Simplify

u = 3 u - 1

u = 3 u - 1

Move

3 u

to the left side

u = 3 u - 1

Subtract

3 u

from both sides

u - 3 u = 3 u - 1 - 3 u

Simplify

- 2 u = - 1

- 2 u = - 1

Divide both sides by

- 2

- 2 u = - 1

Divide both sides by

- 2

\frac{- 2 u}{- 2} = \frac{- 1}{- 2}

Simplify

u = \frac{1}{2}

u = \frac{1}{2}

Substitute back

u = sin (θ)

sin (θ) = \frac{1}{2}

sin (θ) = \frac{1}{2}

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (θ) = \frac{1}{2}

General solutions for

sin (θ) = \frac{1}{2}

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Combine all the solutions

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin^{2} (x) - \frac{3}{4} = 0

tan (x) = - 8

- 3 sin (2 x) = 0

sin (x) = 0.35

tan (x) = \frac{- 1}{3}

What is the general solution for 1+sin(θ)=3sin(θ) ?
The general solution for 1+sin(θ)=3sin(θ) is θ= pi/6+2pin,θ=(5pi)/6+2pin