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

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

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sin^3(x)=-1

Solution

sin^{3} (x) = - 1

Solution

x = \frac{3 π}{2} + 2 πn

Degrees

x = 27 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

n

is odd

= - 1

Simplify

n

is odd

= - 1

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

Multiply:

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

Remove parentheses:

(a) = a, - (- a) = a

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

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

Simplify

n

is odd

= - 1

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

Multiply:

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

Apply rule

- (- a) = a

= \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{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

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{3 π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

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