What is the general solution for 3sin(θ)=sin(θ)-sqrt(2) ?

The general solution for 3sin(θ)=sin(θ)-sqrt(2) is θ=(5pi)/4+2pin,θ=(7pi)/4+2pin

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3sin(θ)=sin(θ)-sqrt(2)

3 sin (θ) = sin (θ) - 2

θ = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

Degrees

θ = 22 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n

Solution steps

3 sin (θ) = sin (θ) - 2

Solve by substitution

3 sin (θ) = sin (θ) - 2

Let:

sin (θ) = u

3 u = u - 2

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

3 u = u - 2

Move

u

to the left side

3 u = u - 2

Subtract

u

from both sides

3 u - u = u - 2 - u

Simplify

2 u = - 2

2 u = - 2

Divide both sides by

2

2 u = - 2

Divide both sides by

2

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

Simplify

u = - \frac{2}{2}

u = - \frac{2}{2}

Substitute back

u = sin (θ)

sin (θ) = - \frac{2}{2}

sin (θ) = - \frac{2}{2}

sin (θ) = - \frac{2}{2} : θ = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

sin (θ) = - \frac{2}{2}

General solutions for

sin (θ) = - \frac{2}{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{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

θ = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

Combine all the solutions

θ = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

What is the general solution for 3sin(θ)=sin(θ)-sqrt(2) ?
The general solution for 3sin(θ)=sin(θ)-sqrt(2) is θ=(5pi)/4+2pin,θ=(7pi)/4+2pin