What is the general solution for 2sin(pi/3-x)=1 ?

The general solution for 2sin(pi/3-x)=1 is x=-2pin+pi/6 ,x=-2pin-pi/2

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2sin(pi/3-x)=1

Solution

2 sin (\frac{π}{3} - x) = 1

Solution

x = - 2 πn + \frac{π}{6}, x = - 2 πn - \frac{π}{2}

Degrees

x = 3 0^{\circ} - 36 0^{\circ} n, x = - 9 0^{\circ} - 36 0^{\circ} n

Solution steps

2 sin (\frac{π}{3} - x) = 1

Divide both sides by

2

2 sin (\frac{π}{3} - x) = 1

Divide both sides by

2

\frac{2 sin ( \frac{π}{3} - x )}{2} = \frac{1}{2}

Simplify

sin (\frac{π}{3} - x) = \frac{1}{2}

sin (\frac{π}{3} - x) = \frac{1}{2}

General solutions for

sin (\frac{π}{3} - x) = \frac{1}{2}

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}

\frac{π}{3} - x = \frac{π}{6} + 2 πn, \frac{π}{3} - x = \frac{5 π}{6} + 2 πn

\frac{π}{3} - x = \frac{π}{6} + 2 πn, \frac{π}{3} - x = \frac{5 π}{6} + 2 πn

Solve

\frac{π}{3} - x = \frac{π}{6} + 2 πn : x = - 2 πn + \frac{π}{6}

\frac{π}{3} - x = \frac{π}{6} + 2 πn

Move

\frac{π}{3}

to the right side

\frac{π}{3} - x = \frac{π}{6} + 2 πn

Subtract

\frac{π}{3}

from both sides

\frac{π}{3} - x - \frac{π}{3} = \frac{π}{6} + 2 πn - \frac{π}{3}

Simplify

\frac{π}{3} - x - \frac{π}{3} = \frac{π}{6} + 2 πn - \frac{π}{3}

Simplify

\frac{π}{3} - x - \frac{π}{3} : - x

\frac{π}{3} - x - \frac{π}{3}

Add similar elements:

\frac{π}{3} - \frac{π}{3} = 0

= - x

Simplify

\frac{π}{6} + 2 πn - \frac{π}{3} : 2 πn - \frac{π}{6}

\frac{π}{6} + 2 πn - \frac{π}{3}

Group like terms

= 2 πn + \frac{π}{6} - \frac{π}{3}

Least Common Multiplier of

6, 3 : 6

6, 3

Least Common Multiplier (LCM)

Prime factorization of

6 : 2 \cdot 3

6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

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

6

3

= 2 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 6

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

6

For

\frac{π}{3} :

multiply the denominator and numerator by

2

\frac{π}{3} = \frac{π 2}{3 \cdot 2} = \frac{π 2}{6}

= \frac{π}{6} - \frac{π 2}{6}

Since the denominators are equal, combine the fractions:

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

= \frac{π - π 2}{6}

Add similar elements:

π - 2 π = - π

= \frac{- π}{6}

Apply the fraction rule:

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

= 2 πn - \frac{π}{6}

- x = 2 πn - \frac{π}{6}

- x = 2 πn - \frac{π}{6}

- x = 2 πn - \frac{π}{6}

Divide both sides by

- 1

- x = 2 πn - \frac{π}{6}

Divide both sides by

- 1

\frac{- x}{- 1} = \frac{2 πn}{- 1} - \frac{\frac{π}{6}}{- 1}

Simplify

\frac{- x}{- 1} = \frac{2 πn}{- 1} - \frac{\frac{π}{6}}{- 1}

Simplify

\frac{- x}{- 1} : x

\frac{- x}{- 1}

Apply the fraction rule:

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

= \frac{x}{1}

Apply rule

\frac{a}{1} = a

= x

Simplify

\frac{2 πn}{- 1} - \frac{\frac{π}{6}}{- 1} : - 2 πn + \frac{π}{6}

\frac{2 πn}{- 1} - \frac{\frac{π}{6}}{- 1}

\frac{2 πn}{- 1} = - 2 πn

\frac{2 πn}{- 1}

Apply the fraction rule:

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

= - \frac{2 πn}{1}

Apply rule

\frac{a}{1} = a

= - 2 πn

= - 2 πn - \frac{\frac{π}{6}}{- 1}

\frac{\frac{π}{6}}{- 1} = - \frac{π}{6}

\frac{\frac{π}{6}}{- 1}

Apply the fraction rule:

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

= - \frac{\frac{π}{6}}{1}

Apply the fraction rule:

\frac{a}{1} = a

\frac{\frac{π}{6}}{1} = \frac{π}{6}

= - \frac{π}{6}

= - 2 πn - (- \frac{π}{6})

Apply rule

- (- a) = a

= - 2 πn + \frac{π}{6}

x = - 2 πn + \frac{π}{6}

x = - 2 πn + \frac{π}{6}

x = - 2 πn + \frac{π}{6}

Solve

\frac{π}{3} - x = \frac{5 π}{6} + 2 πn : x = - 2 πn - \frac{π}{2}

\frac{π}{3} - x = \frac{5 π}{6} + 2 πn

Move

\frac{π}{3}

to the right side

\frac{π}{3} - x = \frac{5 π}{6} + 2 πn

Subtract

\frac{π}{3}

from both sides

\frac{π}{3} - x - \frac{π}{3} = \frac{5 π}{6} + 2 πn - \frac{π}{3}

Simplify

\frac{π}{3} - x - \frac{π}{3} = \frac{5 π}{6} + 2 πn - \frac{π}{3}

Simplify

\frac{π}{3} - x - \frac{π}{3} : - x

\frac{π}{3} - x - \frac{π}{3}

Add similar elements:

\frac{π}{3} - \frac{π}{3} = 0

= - x

Simplify

\frac{5 π}{6} + 2 πn - \frac{π}{3} : 2 πn + \frac{π}{2}

\frac{5 π}{6} + 2 πn - \frac{π}{3}

Group like terms

= 2 πn - \frac{π}{3} + \frac{5 π}{6}

Least Common Multiplier of

3, 6 : 6

3, 6

Least Common Multiplier (LCM)

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Prime factorization of

6 : 2 \cdot 3

6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3

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

3

6

= 3 \cdot 2

Multiply the numbers:

3 \cdot 2 = 6

= 6

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

6

For

\frac{π}{3} :

multiply the denominator and numerator by

2

\frac{π}{3} = \frac{π 2}{3 \cdot 2} = \frac{π 2}{6}

= - \frac{π 2}{6} + \frac{5 π}{6}

Since the denominators are equal, combine the fractions:

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

= \frac{- π 2 + 5 π}{6}

Add similar elements:

- 2 π + 5 π = 3 π

= \frac{3 π}{6}

Cancel the common factor:

3

= 2 πn + \frac{π}{2}

- x = 2 πn + \frac{π}{2}

- x = 2 πn + \frac{π}{2}

- x = 2 πn + \frac{π}{2}

Divide both sides by

- 1

- x = 2 πn + \frac{π}{2}

Divide both sides by

- 1

\frac{- x}{- 1} = \frac{2 πn}{- 1} + \frac{\frac{π}{2}}{- 1}

Simplify

\frac{- x}{- 1} = \frac{2 πn}{- 1} + \frac{\frac{π}{2}}{- 1}

Simplify

\frac{- x}{- 1} : x

\frac{- x}{- 1}

Apply the fraction rule:

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

= \frac{x}{1}

Apply rule

\frac{a}{1} = a

= x

Simplify

\frac{2 πn}{- 1} + \frac{\frac{π}{2}}{- 1} : - 2 πn - \frac{π}{2}

\frac{2 πn}{- 1} + \frac{\frac{π}{2}}{- 1}

\frac{2 πn}{- 1} = - 2 πn

\frac{2 πn}{- 1}

Apply the fraction rule:

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

= - \frac{2 πn}{1}

Apply rule

\frac{a}{1} = a

= - 2 πn

= - 2 πn + \frac{\frac{π}{2}}{- 1}

\frac{\frac{π}{2}}{- 1} = - \frac{π}{2}

\frac{\frac{π}{2}}{- 1}

Apply the fraction rule:

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

= - \frac{\frac{π}{2}}{1}

Apply the fraction rule:

\frac{a}{1} = a

\frac{\frac{π}{2}}{1} = \frac{π}{2}

= - \frac{π}{2}

= - 2 πn - \frac{π}{2}

x = - 2 πn - \frac{π}{2}

x = - 2 πn - \frac{π}{2}

x = - 2 πn - \frac{π}{2}

x = - 2 πn + \frac{π}{6}, x = - 2 πn - \frac{π}{2}

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin(pi/3-x)=1 ?
The general solution for 2sin(pi/3-x)=1 is x=-2pin+pi/6 ,x=-2pin-pi/2