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

Q: What is the general solution for sin^3(x)+3sin^2(x)+2sin(x)=0,0<= x<2pi ?

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

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sin^3(x)+3sin^2(x)+2sin(x)=0,0<= x<2pi

Solution

sin^{3} (x) + 3 sin^{2} (x) + 2 sin (x) = 0, 0 \leq x < 2 π

Solution

x = 0, x = π, x = \frac{3 π}{2}

Degrees

x = 0^{\circ}, x = 18 0^{\circ}, x = 27 0^{\circ}

Solution steps

sin^{3} (x) + 3 sin^{2} (x) + 2 sin (x) = 0, 0 \leq x < 2 π

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Factor

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

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

Factor out common term

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

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

Apply exponent rule:

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

u^{2} = uu

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

Factor out common term

u

= u (u^{2} + 3 u + 2)

= u (u^{2} + 3 u + 2)

Factor

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

u^{2} + 3 u + 2

Break the expression into groups

u^{2} + 3 u + 2

Definition

Factors of

2 : 1, 2

2

Divisors (Factors)

Find the Prime factors of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Add 1

1

The factors of

2

1, 2

For every two factors such that

u * v = 2,

check if

u + v = 3

Check

u = 1, v = 2 : u * v = 2, u + v = 3 \Rightarrow

True

u = 1, v = 2

Group into

(a x^{2} + ux) + (vx + c)

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

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

Factor out

u

from

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

u^{2} + u

Apply exponent rule:

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

u^{2} = uu

= uu + u

Factor out common term

u

= u (u + 1)

Factor out

2

from

2 u + 2 : 2 (u + 1)

2 u + 2

Factor out common term

2

= 2 (u + 1)

= u (u + 1) + 2 (u + 1)

Factor out common term

u + 1

= (u + 1) (u + 2)

= u (u + 1) (u + 2)

u (u + 1) (u + 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u = 0 or u + 1 = 0 or u + 2 = 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 + 2 = 0 : u = - 2

u + 2 = 0

Move

2

to the right side

u + 2 = 0

Subtract

2

from both sides

u + 2 - 2 = 0 - 2

Simplify

u = - 2

u = - 2

The solutions are

u = 0, u = - 1, u = - 2

Substitute back

u = sin (x)

sin (x) = 0, sin (x) = - 1, sin (x) = - 2

sin (x) = 0, sin (x) = - 1, sin (x) = - 2

sin (x) = 0, 0 \leq x < 2 π : x = 0, x = π

sin (x) = 0, 0 \leq x < 2 π

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

Solutions for the range

0 \leq x < 2 π

x = 0, x = π

sin (x) = - 1, 0 \leq x < 2 π : x = \frac{3 π}{2}

sin (x) = - 1, 0 \leq x < 2 π

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

Solutions for the range

0 \leq x < 2 π

x = \frac{3 π}{2}

sin (x) = - 2, 0 \leq x < 2 π :

No Solution

sin (x) = - 2, 0 \leq x < 2 π

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = 0, x = π, x = \frac{3 π}{2}

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Frequently Asked Questions (FAQ)

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