English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Trigonometry >

solvefor x,b*f=sin^3(x)

Solution

solvefor

Solution

x = arcsin (3 b f) + 2 πn, x = π + arcsin (- 3 b f) + 2 πn

Solution steps

b f = sin^{3} (x)

Switch sides

sin^{3} (x) = b f

Solve by substitution

sin^{3} (x) = b f

Let:

sin (x) = u

u^{3} = b f

u^{3} = b f : u = 3 b f, u = - \frac{3 b f}{2} + i \frac{3 3 b f}{2}, u = - \frac{3 b f}{2} - i \frac{3 3 b f}{2}

u^{3} = b f

For

x^{3} = f (a)

the solutions are

x = 3 f (a), 3 f (a) \frac{- 1 - 3 i}{2}, 3 f (a) \frac{- 1 + 3 i}{2}

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

Simplify

3 b f \frac{- 1 + 3 i}{2} : - \frac{3 b f}{2} + i \frac{3 3 b f}{2}

3 b f \frac{- 1 + 3 i}{2}

Rewrite

3 b f \frac{- 1 + 3 i}{2}

in standard complex form:

- \frac{3 b f}{2} + \frac{3 3 b f}{2} i

3 b f \frac{- 1 + 3 i}{2}

Multiply fractions:

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

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

Expand

(- 1 + 3 i) 3 b f : - 3 b f + 3 i 3 b f

(- 1 + 3 i) 3 b f

Apply the distributive law:

a (b + c) = ab + a c

a = 3 b f, b = - 1, c = 3 i

= 3 b f (- 1) + 3 b f 3 i

Apply minus-plus rules

+ (- a) = - a

= - 1 \cdot 3 b f + 3 i 3 b f

Multiply:

1 \cdot 3 b f = 3 b f

= - 3 b f + 3 i 3 b f

= \frac{- 3 b f + 3 i 3 b f}{2}

Apply the fraction rule:

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

\frac{- 3 b f + 3 i 3 b f}{2} = - \frac{3 b f}{2} + \frac{3 i 3 b f}{2}

= - \frac{3 b f}{2} + \frac{3 3 b f}{2}

= - \frac{3 b f}{2} + \frac{3 3 b f}{2} i

Simplify

3 b f \frac{- 1 - 3 i}{2} : - \frac{3 b f}{2} - i \frac{3 3 b f}{2}

3 b f \frac{- 1 - 3 i}{2}

Rewrite

3 b f \frac{- 1 - 3 i}{2}

in standard complex form:

- \frac{3 b f}{2} - \frac{3 3 b f}{2} i

3 b f \frac{- 1 - 3 i}{2}

Multiply fractions:

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

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

Expand

(- 1 - 3 i) 3 b f : - 3 b f - 3 i 3 b f

(- 1 - 3 i) 3 b f

Apply the distributive law:

a (b - c) = ab - a c

a = 3 b f, b = - 1, c = 3 i

= 3 b f (- 1) - 3 b f 3 i

Apply minus-plus rules

+ (- a) = - a

= - 1 \cdot 3 b f - 3 i 3 b f

Multiply:

1 \cdot 3 b f = 3 b f

= - 3 b f - 3 i 3 b f

= \frac{- 3 b f - 3 i 3 b f}{2}

Apply the fraction rule:

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

\frac{- 3 b f - 3 i 3 b f}{2} = - \frac{3 b f}{2} - \frac{3 i 3 b f}{2}

= - \frac{3 b f}{2} - \frac{3 3 b f}{2}

= - \frac{3 b f}{2} - \frac{3 3 b f}{2} i

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

Substitute back

u = sin (x)

sin (x) = 3 b f, sin (x) = - \frac{3 b f}{2} + i \frac{3 3 b f}{2}, sin (x) = - \frac{3 b f}{2} - i \frac{3 3 b f}{2}

sin (x) = 3 b f, sin (x) = - \frac{3 b f}{2} + i \frac{3 3 b f}{2}, sin (x) = - \frac{3 b f}{2} - i \frac{3 3 b f}{2}

sin (x) = 3 b f : x = arcsin (3 b f) + 2 πn, x = π + arcsin (- 3 b f) + 2 πn

sin (x) = 3 b f

Apply trig inverse properties

sin (x) = 3 b f

General solutions for

sin (x) = 3 b f

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (3 b f) + 2 πn, x = π + arcsin (- 3 b f) + 2 πn

x = arcsin (3 b f) + 2 πn, x = π + arcsin (- 3 b f) + 2 πn

sin (x) = - \frac{3 b f}{2} + i \frac{3 3 b f}{2} :

No Solution

sin (x) = - \frac{3 b f}{2} + i \frac{3 3 b f}{2}

No Solution

sin (x) = - \frac{3 b f}{2} - i \frac{3 3 b f}{2} :

No Solution

sin (x) = - \frac{3 b f}{2} - i \frac{3 3 b f}{2}

No Solution

Combine all the solutions

x = arcsin (3 b f) + 2 πn, x = π + arcsin (- 3 b f) + 2 πn

Graph

Popular Examples

2sin^2(x)+sin^3(x)-1=0

2cos^2(x)=3cos(x)-1

sin^2(x)-4sin(x)+4=0

3cos^2(x)-10cos(x)+3=0

(m+1)sin(x)+2-m=0