What is the general solution for sin^5(x)+sin^3(x)=0 ?

The general solution for sin^5(x)+sin^3(x)=0 is x=2pin,x=pi+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin^5(x)+sin^3(x)=0

Solution

sin^{5} (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^{5} (x) + sin^{3} (x) = 0

Solve by substitution

sin^{5} (x) + sin^{3} (x) = 0

Let:

sin (x) = u

u^{5} + u^{3} = 0

u^{5} + u^{3} = 0 : u = 0, u = i, u = - i

u^{5} + u^{3} = 0

Factor

u^{5} + u^{3} : u^{3} (u^{2} + 1)

u^{5} + u^{3}

Apply exponent rule:

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

u^{5} = u^{2} u^{3}

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

Factor out common term

u^{3}

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

u^{3} (u^{2} + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

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

u^{2} + 1 = 0

Move

1

to the right side

u^{2} + 1 = 0

Subtract

1

from both sides

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

Simplify

u^{2} = - 1

u^{2} = - 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = - 1, u = - - 1

Simplify

- 1 : i

- 1

Apply imaginary number rule:

- 1 = i

= i

Simplify

- - 1 : - i

- - 1

Apply imaginary number rule:

- 1 = i

= - i

u = i, u = - i

The solutions are

u = 0, u = i, u = - i

Substitute back

u = sin (x)

sin (x) = 0, sin (x) = i, sin (x) = - i

sin (x) = 0, sin (x) = i, sin (x) = - i

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) = i :

No Solution

sin (x) = i

No Solution

sin (x) = - i :

No Solution

sin (x) = - i

No Solution

Combine all the solutions

x = 2 πn, x = π + 2 πn

Popular Examples

5sin^2(x)cos(7x)-cos(7x)=0

5 sin^{2} (x) cos (7 x) - cos (7 x) = 0

solvefor x,cos^2(x)*z^2-2cos^2(x)*z+1=0 solve for

x, cos^{2} (x) \cdot z^{2} - 2 cos^{2} (x) \cdot z + 1 = 0

sin(a)+sin(120+a)+sin(120-a)=0

sin (a) + sin (12 0^{\circ} + a) + sin (12 0^{\circ} - a) = 0

3sin(x)sin(x)=5cos(x)-2

3 sin (x) sin (x) = 5 cos (x) - 2

(cos(x)+3cos(x))/(2+2)=0

\frac{cos ( x ) + 3 cos ( x )}{2 + 2} = 0

Frequently Asked Questions (FAQ)

What is the general solution for sin^5(x)+sin^3(x)=0 ?
The general solution for sin^5(x)+sin^3(x)=0 is x=2pin,x=pi+2pin