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

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

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sin^3(t)=sin(t)

Solution

sin^{3} (t) = sin (t)

Solution

t = 2 πn, t = π + 2 πn, t = \frac{3 π}{2} + 2 πn, t = \frac{π}{2} + 2 πn

Degrees

t = 0^{\circ} + 36 0^{\circ} n, t = 18 0^{\circ} + 36 0^{\circ} n, t = 27 0^{\circ} + 36 0^{\circ} n, t = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

sin^{3} (t) = sin (t)

Solve by substitution

sin^{3} (t) = sin (t)

Let:

sin (t) = u

u^{3} = u

u^{3} = u : u = 0, u = - 1, u = 1

u^{3} = u

Move

u

to the left side

u^{3} = u

Subtract

u

from both sides

u^{3} - u = u - u

Simplify

u^{3} - u = 0

u^{3} - u = 0

Factor

u^{3} - u : u (u + 1) (u - 1)

u^{3} - u

Factor out common term

u : u (u^{2} - 1)

u^{3} - u

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

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

= u^{2} u - u

Factor out common term

u

= u (u^{2} - 1)

= u (u^{2} - 1)

Factor

u^{2} - 1 : (u + 1) (u - 1)

u^{2} - 1

Rewrite

1

1^{2}

= u^{2} - 1^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

u^{2} - 1^{2} = (u + 1) (u - 1)

= (u + 1) (u - 1)

= u (u + 1) (u - 1)

u (u + 1) (u - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u = 0 or u + 1 = 0 or u - 1 = 0

Solve

u + 1 = 0 : u = - 1

u + 1 = 0

Move

1

to the right side

u + 1 = 0

Subtract

1

from both sides

u + 1 - 1 = 0 - 1

Simplify

u = - 1

u = - 1

Solve

u - 1 = 0 : u = 1

u - 1 = 0

Move

1

to the right side

u - 1 = 0

Add

1

to both sides

u - 1 + 1 = 0 + 1

Simplify

u = 1

u = 1

The solutions are

u = 0, u = - 1, u = 1

Substitute back

u = sin (t)

sin (t) = 0, sin (t) = - 1, sin (t) = 1

sin (t) = 0, sin (t) = - 1, sin (t) = 1

sin (t) = 0 : t = 2 πn, t = π + 2 πn

sin (t) = 0

General solutions for

sin (t) = 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}

t = 0 + 2 πn, t = π + 2 πn

t = 0 + 2 πn, t = π + 2 πn

Solve

t = 0 + 2 πn : t = 2 πn

t = 0 + 2 πn

0 + 2 πn = 2 πn

t = 2 πn

t = 2 πn, t = π + 2 πn

sin (t) = - 1 : t = \frac{3 π}{2} + 2 πn

sin (t) = - 1

General solutions for

sin (t) = - 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}

t = \frac{3 π}{2} + 2 πn

t = \frac{3 π}{2} + 2 πn

sin (t) = 1 : t = \frac{π}{2} + 2 πn

sin (t) = 1

General solutions for

sin (t) = 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}

t = \frac{π}{2} + 2 πn

t = \frac{π}{2} + 2 πn

Combine all the solutions

t = 2 πn, t = π + 2 πn, t = \frac{3 π}{2} + 2 πn, t = \frac{π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

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