What is the general solution for 2sin^3(x)=sin^3(x) ?

The general solution for 2sin^3(x)=sin^3(x) is x=2pin,x=pi+2pin

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2sin^3(x)=sin^3(x)

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

Solve by substitution

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

Let:

sin (x) = u

2 u^{3} = u^{3}

2 u^{3} = u^{3} : u = 0

2 u^{3} = u^{3}

Move

u^{3}

to the left side

2 u^{3} = u^{3}

Subtract

u^{3}

from both sides

2 u^{3} - u^{3} = u^{3} - u^{3}

Simplify

u^{3} = 0

u^{3} = 0

Apply rule

x^{n} = 0 \Rightarrow x = 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

θ, y = 4 sin (θ)

- 3 (sin^{2} (θ) - cos^{2} (θ)) = 0

cos^{2} (t) - sin^{2} (t) = - 1

- a^{2} c_{1} cos (a x) - a^{2} c_{2} sin (a x) = 0

10 = - 12 sin (x) + 1.8 cos (x)

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