What is the general solution for sin^3(x)=1 ?

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

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sin^3(x)=1

sin^{3} (x) = 1

x = \frac{π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

sin^{3} (x) = 1

Solve by substitution

sin^{3} (x) = 1

Let:

sin (x) = u

u^{3} = 1

u^{3} = 1 : u = 1, u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

u^{3} = 1

For

x^{3} = f (a)

the solutions are

u = 1, u = \frac{- 1 + 3 i}{2}, u = \frac{- 1 - 3 i}{2}

Simplify

\frac{- 1 + 3 i}{2} : - \frac{1}{2} + i \frac{3}{2}

\frac{- 1 + 3 i}{2}

Rewrite

\frac{- 1 + 3 i}{2}

in standard complex form:

- \frac{1}{2} + \frac{3}{2} i

\frac{- 1 + 3 i}{2}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 1 + 3 i}{2} = - \frac{1}{2} + \frac{3 i}{2}

= - \frac{1}{2} + \frac{3 i}{2}

= - \frac{1}{2} + \frac{3}{2} i

Simplify

\frac{- 1 - 3 i}{2} : - \frac{1}{2} - i \frac{3}{2}

\frac{- 1 - 3 i}{2}

Rewrite

\frac{- 1 - 3 i}{2}

in standard complex form:

- \frac{1}{2} - \frac{3}{2} i

\frac{- 1 - 3 i}{2}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 1 - 3 i}{2} = - \frac{1}{2} - \frac{3 i}{2}

= - \frac{1}{2} - \frac{3 i}{2}

= - \frac{1}{2} - \frac{3}{2} i

u = 1, u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

Substitute back

u = sin (x)

sin (x) = 1, sin (x) = - \frac{1}{2} + i \frac{3}{2}, sin (x) = - \frac{1}{2} - i \frac{3}{2}

sin (x) = 1, sin (x) = - \frac{1}{2} + i \frac{3}{2}, sin (x) = - \frac{1}{2} - i \frac{3}{2}

sin (x) = 1 : x = \frac{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{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

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

No Solution

sin (x) = - \frac{1}{2} + i \frac{3}{2}

No Solution

sin (x) = - \frac{1}{2} - i \frac{3}{2} :

No Solution

sin (x) = - \frac{1}{2} - i \frac{3}{2}

No Solution

Combine all the solutions

x = \frac{π}{2} + 2 πn

cosh (x) = \frac{5}{4}

1 = cos (t)

cos (2 x) + 2 = sin (x)

tan (t) = \frac{8}{15}

3 cos^{2} (x) - 6 cos (x) = - 3

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