What is the general solution for sin(x)+sin^2(x)+sin^3(x)=0 ?

The general solution for sin(x)+sin^2(x)+sin^3(x)=0 is x=2pin,x=pi+2pin

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sin(x)+sin^2(x)+sin^3(x)=0

Solution

sin (x) + sin^{2} (x) + sin^{3} (x) = 0

Solution

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

sin (x) + sin^{2} (x) + sin^{3} (x) = 0

Solve by substitution

sin (x) + sin^{2} (x) + sin^{3} (x) = 0

Let:

sin (x) = u

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

u + u^{2} + u^{3} = 0 : u = 0, u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

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

Factor

u + u^{2} + u^{3} : u (u^{2} + u + 1)

u + u^{2} + u^{3}

Apply exponent rule:

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

u^{2} = uu

= u^{2} u + uu + u

Factor out common term

u

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

u (u^{2} + u + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u = 0 or u^{2} + u + 1 = 0

Solve

u^{2} + u + 1 = 0 : u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

u^{2} + u + 1 = 0

Solve with the quadratic formula

u^{2} + u + 1 = 0

Quadratic Equation Formula:

For

a = 1, b = 1, c = 1

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

Simplify

1^{2} - 4 \cdot 1 \cdot 1 : 3 i

1^{2} - 4 \cdot 1 \cdot 1

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 1 - 4

Subtract the numbers:

1 - 4 = - 3

= - 3

Apply radical rule:

- a = - 1 a

- 3 = - 1 3

= - 1 3

Apply imaginary number rule:

- 1 = i

= 3 i

u_{1, 2} = \frac{- 1 \pm 3 i}{2 \cdot 1}

Separate the solutions

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

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

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

Multiply the numbers:

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

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

Multiply the numbers:

2 \cdot 1 = 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

The solutions to the quadratic equation are:

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

The solutions are

u = 0, u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

Substitute back

u = sin (x)

sin (x) = 0, sin (x) = - \frac{1}{2} + i \frac{3}{2}, sin (x) = - \frac{1}{2} - i \frac{3}{2}

sin (x) = 0, sin (x) = - \frac{1}{2} + i \frac{3}{2}, sin (x) = - \frac{1}{2} - i \frac{3}{2}

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

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 = 2 πn, x = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for sin(x)+sin^2(x)+sin^3(x)=0 ?
The general solution for sin(x)+sin^2(x)+sin^3(x)=0 is x=2pin,x=pi+2pin