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

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

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sin(x)=2sin(x)+1

sin (x) = 2 sin (x) + 1

x = \frac{3 π}{2} + 2 πn

Degrees

x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (x) = 2 sin (x) + 1

Solve by substitution

sin (x) = 2 sin (x) + 1

Let:

sin (x) = 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 (x)

sin (x) = - 1

sin (x) = - 1

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

sin (x) = - 1

General solutions for

sin (x) = - 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}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{3 π}{2} + 2 πn

4 cosh (x) - sinh (x) = 8

- 53 cot (θ) + 1 = 16

- sin (x) = sin (x)

2 sin^{2} (x) + 4 sin (x) + 2 = 0

2 sin (θ) = csc (θ)

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