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

The general solution for 2sin(x)=sin(x) is x=2pin,x=pi+2pin

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2sin(x)=sin(x)

2 sin (x) = sin (x)

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

2 sin (x) = sin (x)

Solve by substitution

2 sin (x) = sin (x)

Let:

sin (x) = u

2 u = u

2 u = u : u = 0

2 u = u

Move

u

to the left side

2 u = u

Subtract

u

from both sides

2 u - u = u - u

Simplify

u = 0

u = 0

Substitute back

u = sin (x)

sin (x) = 0

sin (x) = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn

- sin (2 θ) - cos (4 θ) = 0

5 tan (θ) + 9 = 0

3 sin (θ) = 3 sin (θ) + 3 cos (θ) + 3 sin (θ)

2 sin^{2} (x) - 3 sin (x) = 2

tan (x + π) + 2 sin (x + π) = 0, 0 \leq x \leq 2 π

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