What is the general solution for -4+sin(-3x+(2pi)/3)=-4 ?

The general solution for -4+sin(-3x+(2pi)/3)=-4 is x=-(2pin)/3+(2pi)/9 ,x=-pi/3+(2pi)/9-(2pin)/3

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Popular >

-4+sin(-3x+(2pi)/3)=-4

Solution

- 4 + sin (- 3 x + \frac{2 π}{3}) = - 4

Solution

x = - \frac{2 πn}{3} + \frac{2 π}{9}, x = - \frac{π}{3} + \frac{2 π}{9} - \frac{2 πn}{3}

Degrees

x = 4 0^{\circ} - 12 0^{\circ} n, x = - 2 0^{\circ} - 12 0^{\circ} n

Solution steps

- 4 + sin (- 3 x + \frac{2 π}{3}) = - 4

Add

4

to both sides

- 4 + sin (- 3 x + \frac{2 π}{3}) + 4 = - 4 + 4

Simplify

sin (- 3 x + \frac{2 π}{3}) = 0

General solutions for

sin (- 3 x + \frac{2 π}{3}) = 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}

- 3 x + \frac{2 π}{3} = 0 + 2 πn, - 3 x + \frac{2 π}{3} = π + 2 πn

- 3 x + \frac{2 π}{3} = 0 + 2 πn, - 3 x + \frac{2 π}{3} = π + 2 πn

Solve

- 3 x + \frac{2 π}{3} = 0 + 2 πn : x = - \frac{2 πn}{3} + \frac{2 π}{9}

- 3 x + \frac{2 π}{3} = 0 + 2 πn

0 + 2 πn = 2 πn

- 3 x + \frac{2 π}{3} = 2 πn

Move

\frac{2 π}{3}

to the right side

- 3 x + \frac{2 π}{3} = 2 πn

Subtract

\frac{2 π}{3}

from both sides

- 3 x + \frac{2 π}{3} - \frac{2 π}{3} = 2 πn - \frac{2 π}{3}

Simplify

- 3 x = 2 πn - \frac{2 π}{3}

- 3 x = 2 πn - \frac{2 π}{3}

Divide both sides by

- 3

- 3 x = 2 πn - \frac{2 π}{3}

Divide both sides by

- 3

\frac{- 3 x}{- 3} = \frac{2 πn}{- 3} - \frac{\frac{2 π}{3}}{- 3}

Simplify

\frac{- 3 x}{- 3} = \frac{2 πn}{- 3} - \frac{\frac{2 π}{3}}{- 3}

Simplify

\frac{- 3 x}{- 3} : x

\frac{- 3 x}{- 3}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{2 πn}{- 3} - \frac{\frac{2 π}{3}}{- 3} : - \frac{2 πn}{3} + \frac{2 π}{9}

\frac{2 πn}{- 3} - \frac{\frac{2 π}{3}}{- 3}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{2 πn}{3} - \frac{\frac{2 π}{3}}{- 3}

\frac{\frac{2 π}{3}}{- 3} = - \frac{2 π}{9}

\frac{\frac{2 π}{3}}{- 3}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{\frac{2 π}{3}}{3}

Apply the fraction rule:

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

\frac{\frac{2 π}{3}}{3} = \frac{2 π}{3 \cdot 3}

= - \frac{2 π}{3 \cdot 3}

Multiply the numbers:

3 \cdot 3 = 9

= - \frac{2 π}{9}

= - \frac{2 πn}{3} - (- \frac{2 π}{9})

Apply rule

- (- a) = a

= - \frac{2 πn}{3} + \frac{2 π}{9}

x = - \frac{2 πn}{3} + \frac{2 π}{9}

x = - \frac{2 πn}{3} + \frac{2 π}{9}

x = - \frac{2 πn}{3} + \frac{2 π}{9}

Solve

- 3 x + \frac{2 π}{3} = π + 2 πn : x = - \frac{π}{3} + \frac{2 π}{9} - \frac{2 πn}{3}

- 3 x + \frac{2 π}{3} = π + 2 πn

Move

\frac{2 π}{3}

to the right side

- 3 x + \frac{2 π}{3} = π + 2 πn

Subtract

\frac{2 π}{3}

from both sides

- 3 x + \frac{2 π}{3} - \frac{2 π}{3} = π + 2 πn - \frac{2 π}{3}

Simplify

- 3 x = π + 2 πn - \frac{2 π}{3}

- 3 x = π + 2 πn - \frac{2 π}{3}

Divide both sides by

- 3

- 3 x = π + 2 πn - \frac{2 π}{3}

Divide both sides by

- 3

\frac{- 3 x}{- 3} = \frac{π}{- 3} + \frac{2 πn}{- 3} - \frac{\frac{2 π}{3}}{- 3}

Simplify

\frac{- 3 x}{- 3} = \frac{π}{- 3} + \frac{2 πn}{- 3} - \frac{\frac{2 π}{3}}{- 3}

Simplify

\frac{- 3 x}{- 3} : x

\frac{- 3 x}{- 3}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{π}{- 3} + \frac{2 πn}{- 3} - \frac{\frac{2 π}{3}}{- 3} : - \frac{π}{3} + \frac{2 π}{9} - \frac{2 πn}{3}

\frac{π}{- 3} + \frac{2 πn}{- 3} - \frac{\frac{2 π}{3}}{- 3}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{π}{3} + \frac{2 πn}{- 3} - \frac{\frac{2 π}{3}}{- 3}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{π}{3} - \frac{2 πn}{3} - \frac{\frac{2 π}{3}}{- 3}

\frac{\frac{2 π}{3}}{- 3} = - \frac{2 π}{9}

\frac{\frac{2 π}{3}}{- 3}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{\frac{2 π}{3}}{3}

Apply the fraction rule:

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

\frac{\frac{2 π}{3}}{3} = \frac{2 π}{3 \cdot 3}

= - \frac{2 π}{3 \cdot 3}

Multiply the numbers:

3 \cdot 3 = 9

= - \frac{2 π}{9}

= - \frac{π}{3} - \frac{2 πn}{3} - (- \frac{2 π}{9})

Apply rule

- (- a) = a

= - \frac{π}{3} - \frac{2 πn}{3} + \frac{2 π}{9}

Group like terms

= - \frac{π}{3} + \frac{2 π}{9} - \frac{2 πn}{3}

x = - \frac{π}{3} + \frac{2 π}{9} - \frac{2 πn}{3}

x = - \frac{π}{3} + \frac{2 π}{9} - \frac{2 πn}{3}

x = - \frac{π}{3} + \frac{2 π}{9} - \frac{2 πn}{3}

x = - \frac{2 πn}{3} + \frac{2 π}{9}, x = - \frac{π}{3} + \frac{2 π}{9} - \frac{2 πn}{3}

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Frequently Asked Questions (FAQ)

What is the general solution for -4+sin(-3x+(2pi)/3)=-4 ?
The general solution for -4+sin(-3x+(2pi)/3)=-4 is x=-(2pin)/3+(2pi)/9 ,x=-pi/3+(2pi)/9-(2pin)/3