What is the general solution for sin(θ)=2sin(θ)+1 ?

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

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sin(θ)=2sin(θ)+1

sin (θ) = 2 sin (θ) + 1

θ = \frac{3 π}{2} + 2 πn

Degrees

θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (θ) = 2 sin (θ) + 1

Solve by substitution

sin (θ) = 2 sin (θ) + 1

Let:

sin (θ) = u

u = 2 u + 1

u = 2 u + 1 : u = - 1

u = 2 u + 1

Move

2 u

to the left side

u = 2 u + 1

Subtract

2 u

from both sides

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

Simplify

- u = 1

- u = 1

Divide both sides by

- 1

- u = 1

Divide both sides by

- 1

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

Simplify

u = - 1

u = - 1

Substitute back

u = sin (θ)

sin (θ) = - 1

sin (θ) = - 1

sin (θ) = - 1 : θ = \frac{3 π}{2} + 2 πn

sin (θ) = - 1

General solutions for

sin (θ) = - 1

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{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

Combine all the solutions

θ = \frac{3 π}{2} + 2 πn

8 sin (x) tan (x) = - 7 tan (x)

cos (x) = \frac{8}{17}

tan (θ) = - \frac{3}{2}

6 cos^{2} (θ) + 13 cos (θ) + 5 = 0

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

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