What is the general solution for sin^3(t)= 1/8 ?

The general solution for sin^3(t)= 1/8 is t= pi/6+2pin,t=(5pi)/6+2pin

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Popular >

sin^3(t)= 1/8

Solution

sin^{3} (t) = \frac{1}{8}

Solution

t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

Degrees

t = 3 0^{\circ} + 36 0^{\circ} n, t = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

sin^{3} (t) = \frac{1}{8}

Solve by substitution

sin^{3} (t) = \frac{1}{8}

Let:

sin (t) = u

u^{3} = \frac{1}{8}

u^{3} = \frac{1}{8} : u = \frac{1}{2}, u = - \frac{1}{4} + i \frac{3}{4}, u = - \frac{1}{4} - i \frac{3}{4}

u^{3} = \frac{1}{8}

For

x^{3} = f (a)

the solutions are

Apply radical rule:

assuming

a \geq 0, b \geq 0

Factor the number:

8 = 2^{3}

Apply radical rule:

= 2

Apply rule

= \frac{1}{2}

Simplify

Apply radical rule:

assuming

a \geq 0, b \geq 0

Factor the number:

8 = 2^{3}

Apply radical rule:

= 2

Apply rule

= \frac{1}{2}

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

Multiply fractions:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

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

1 \cdot (- 1 + 3 i) = - 1 + 3 i

1 \cdot (- 1 + 3 i)

Multiply:

1 \cdot (- 1 + 3 i) = (- 1 + 3 i)

= (- 1 + 3 i)

Remove parentheses:

(- a) = - a

= - 1 + 3 i

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 1 + 3 i}{4}

Rewrite

\frac{- 1 + 3 i}{4}

in standard complex form:

- \frac{1}{4} + \frac{3}{4} i

\frac{- 1 + 3 i}{4}

Apply the fraction rule:

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

\frac{- 1 + 3 i}{4} = - \frac{1}{4} + \frac{3 i}{4}

= - \frac{1}{4} + \frac{3 i}{4}

= - \frac{1}{4} + \frac{3}{4} i

Simplify

Apply radical rule:

assuming

a \geq 0, b \geq 0

Factor the number:

8 = 2^{3}

Apply radical rule:

= 2

Apply rule

= \frac{1}{2}

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

Multiply fractions:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

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

1 \cdot (- 1 - 3 i) = - 1 - 3 i

1 \cdot (- 1 - 3 i)

Multiply:

1 \cdot (- 1 - 3 i) = (- 1 - 3 i)

= (- 1 - 3 i)

Remove parentheses:

(- a) = - a

= - 1 - 3 i

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 1 - 3 i}{4}

Rewrite

\frac{- 1 - 3 i}{4}

in standard complex form:

- \frac{1}{4} - \frac{3}{4} i

\frac{- 1 - 3 i}{4}

Apply the fraction rule:

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

\frac{- 1 - 3 i}{4} = - \frac{1}{4} - \frac{3 i}{4}

= - \frac{1}{4} - \frac{3 i}{4}

= - \frac{1}{4} - \frac{3}{4} i

u = \frac{1}{2}, u = - \frac{1}{4} + i \frac{3}{4}, u = - \frac{1}{4} - i \frac{3}{4}

Substitute back

u = sin (t)

sin (t) = \frac{1}{2}, sin (t) = - \frac{1}{4} + i \frac{3}{4}, sin (t) = - \frac{1}{4} - i \frac{3}{4}

sin (t) = \frac{1}{2}, sin (t) = - \frac{1}{4} + i \frac{3}{4}, sin (t) = - \frac{1}{4} - i \frac{3}{4}

sin (t) = \frac{1}{2} : t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

sin (t) = \frac{1}{2}

General solutions for

sin (t) = \frac{1}{2}

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}

t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

sin (t) = - \frac{1}{4} + i \frac{3}{4} :

No Solution

sin (t) = - \frac{1}{4} + i \frac{3}{4}

No Solution

sin (t) = - \frac{1}{4} - i \frac{3}{4} :

No Solution

sin (t) = - \frac{1}{4} - i \frac{3}{4}

No Solution

Combine all the solutions

t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for sin^3(t)= 1/8 ?
The general solution for sin^3(t)= 1/8 is t= pi/6+2pin,t=(5pi)/6+2pin