What is the general solution for sin(3x-pi/5)=0-1 ?

The general solution for sin(3x-pi/5)=0-1 is x=(2pin)/3+(17pi)/(30)

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sin(3x-pi/5)=0-1

Solution

sin (3 x - \frac{π}{5}) = 0 - 1

Solution

x = \frac{2 πn}{3} + \frac{17 π}{30}

Degrees

x = 10 2^{\circ} + 12 0^{\circ} n

Solution steps

sin (3 x - \frac{π}{5}) = 0 - 1

Subtract the numbers:

0 - 1 = - 1

sin (3 x - \frac{π}{5}) = - 1

General solutions for

sin (3 x - \frac{π}{5}) = - 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}

3 x - \frac{π}{5} = \frac{3 π}{2} + 2 πn

3 x - \frac{π}{5} = \frac{3 π}{2} + 2 πn

Solve

3 x - \frac{π}{5} = \frac{3 π}{2} + 2 πn : x = \frac{2 πn}{3} + \frac{17 π}{30}

3 x - \frac{π}{5} = \frac{3 π}{2} + 2 πn

Move

\frac{π}{5}

to the right side

3 x - \frac{π}{5} = \frac{3 π}{2} + 2 πn

Add

\frac{π}{5}

to both sides

3 x - \frac{π}{5} + \frac{π}{5} = \frac{3 π}{2} + 2 πn + \frac{π}{5}

Simplify

3 x - \frac{π}{5} + \frac{π}{5} = \frac{3 π}{2} + 2 πn + \frac{π}{5}

Simplify

3 x - \frac{π}{5} + \frac{π}{5} : 3 x

3 x - \frac{π}{5} + \frac{π}{5}

Add similar elements:

- \frac{π}{5} + \frac{π}{5} = 0

= 3 x

Simplify

\frac{3 π}{2} + 2 πn + \frac{π}{5} : 2 πn + \frac{17 π}{10}

\frac{3 π}{2} + 2 πn + \frac{π}{5}

Group like terms

= 2 πn + \frac{π}{5} + \frac{3 π}{2}

Least Common Multiplier of

5, 2 : 10

5, 2

Least Common Multiplier (LCM)

Prime factorization of

5 : 5

5

5

is a prime number, therefore no factorization is possible

= 5

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Multiply each factor the greatest number of times it occurs in either

5

2

= 5 \cdot 2

Multiply the numbers:

5 \cdot 2 = 10

= 10

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

10

For

\frac{π}{5} :

multiply the denominator and numerator by

2

\frac{π}{5} = \frac{π 2}{5 \cdot 2} = \frac{π 2}{10}

For

\frac{3 π}{2} :

multiply the denominator and numerator by

5

\frac{3 π}{2} = \frac{3 π 5}{2 \cdot 5} = \frac{15 π}{10}

= \frac{π 2}{10} + \frac{15 π}{10}

Since the denominators are equal, combine the fractions:

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

= \frac{π 2 + 15 π}{10}

Add similar elements:

2 π + 15 π = 17 π

= 2 πn + \frac{17 π}{10}

3 x = 2 πn + \frac{17 π}{10}

3 x = 2 πn + \frac{17 π}{10}

3 x = 2 πn + \frac{17 π}{10}

Divide both sides by

3

3 x = 2 πn + \frac{17 π}{10}

Divide both sides by

3

\frac{3 x}{3} = \frac{2 πn}{3} + \frac{\frac{17 π}{10}}{3}

Simplify

\frac{3 x}{3} = \frac{2 πn}{3} + \frac{\frac{17 π}{10}}{3}

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{2 πn}{3} + \frac{\frac{17 π}{10}}{3} : \frac{2 πn}{3} + \frac{17 π}{30}

\frac{2 πn}{3} + \frac{\frac{17 π}{10}}{3}

\frac{\frac{17 π}{10}}{3} = \frac{17 π}{30}

\frac{\frac{17 π}{10}}{3}

Apply the fraction rule:

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

= \frac{17 π}{10 \cdot 3}

Multiply the numbers:

10 \cdot 3 = 30

= \frac{17 π}{30}

= \frac{2 πn}{3} + \frac{17 π}{30}

x = \frac{2 πn}{3} + \frac{17 π}{30}

x = \frac{2 πn}{3} + \frac{17 π}{30}

x = \frac{2 πn}{3} + \frac{17 π}{30}

x = \frac{2 πn}{3} + \frac{17 π}{30}

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Frequently Asked Questions (FAQ)

What is the general solution for sin(3x-pi/5)=0-1 ?
The general solution for sin(3x-pi/5)=0-1 is x=(2pin)/3+(17pi)/(30)