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

The general solution for sin(x/2)=-cos(x-pi/3) is No Solution for x\in\mathbb{R}

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sin(x/2)=-cos(x-pi/3)

Solution

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

Solution

No Solution for x \in R

Solution steps

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

Multiply by

- 1

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

Rewrite using trig identities

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

Use the following identity:

- sin (x) = sin (- x)

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

Use the following identity:

cos (x) = sin (\frac{π}{2} - x)

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

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

Apply trig inverse properties

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

sin (x) = sin (y) \Rightarrow x = y + 2 π n, x = π - y + 2 π n

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

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

- (\frac{x}{2}) = \frac{π}{2} - (x - \frac{π}{3}) + 2 πn : x = \frac{5 π + 12 πn}{3}

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

Move

(x - \frac{π}{3})

to the left side

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

Add

(x - \frac{π}{3})

to both sides

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

Simplify

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

Simplify

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

- \frac{x}{2} + x - \frac{π}{3}

Convert element to fraction:

x = \frac{x}{1}

= - \frac{x}{2} - \frac{π}{3} + \frac{x}{1}

Least Common Multiplier of

2, 3, 1 : 6

2, 3, 1

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Prime factorization of

1

Compute a number comprised of factors that appear in at least one of the following:

2, 3, 1

= 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{x}{2} :

multiply the denominator and numerator by

3

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

For

\frac{x}{1} :

multiply the denominator and numerator by

6

\frac{x}{1} = \frac{x \cdot 6}{1 \cdot 6} = \frac{x \cdot 6}{6}

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

Since the denominators are equal, combine the fractions:

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

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

- x \cdot 3 - π 2 + x \cdot 6 = 3 x - 2 π

- x \cdot 3 - π 2 + x \cdot 6

Group like terms

= - 3 x + 6 x - 2 π

Add similar elements:

- 3 x + 6 x = 3 x

= 3 x - 2 π

= \frac{3 x - 2 π}{6}

Simplify

\frac{π}{2} - (x - \frac{π}{3}) + 2 πn + x - \frac{π}{3} : \frac{π}{2} + 2 πn

\frac{π}{2} - (x - \frac{π}{3}) + 2 πn + x - \frac{π}{3}

Add similar elements:

- (x - \frac{π}{3}) + x - \frac{π}{3} = 0

= \frac{π}{2} + 2 πn

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

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

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

Multiply both sides by

6

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

Multiply both sides by

6

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

Simplify

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

Simplify

\frac{6 ( 3 x - 2 π )}{6} : 3 x - 2 π

\frac{6 ( 3 x - 2 π )}{6}

Divide the numbers:

\frac{6}{6} = 1

= 3 x - 2 π

Simplify

6 \cdot \frac{π}{2} + 6 \cdot 2 πn : 3 π + 12 πn

6 \cdot \frac{π}{2} + 6 \cdot 2 πn

6 \cdot \frac{π}{2} = 3 π

6 \cdot \frac{π}{2}

Multiply fractions:

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

= \frac{π 6}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3 π

6 \cdot 2 πn = 12 πn

6 \cdot 2 πn

Multiply the numbers:

6 \cdot 2 = 12

= 12 πn

= 3 π + 12 πn

3 x - 2 π = 3 π + 12 πn

3 x - 2 π = 3 π + 12 πn

3 x - 2 π = 3 π + 12 πn

Move

2 π

to the right side

3 x - 2 π = 3 π + 12 πn

Add

2 π

to both sides

3 x - 2 π + 2 π = 3 π + 12 πn + 2 π

Simplify

3 x = 5 π + 12 πn

3 x = 5 π + 12 πn

Divide both sides by

3

3 x = 5 π + 12 πn

Divide both sides by

3

\frac{3 x}{3} = \frac{5 π}{3} + \frac{12 πn}{3}

Simplify

\frac{3 x}{3} = \frac{5 π}{3} + \frac{12 πn}{3}

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{5 π}{3} + \frac{12 πn}{3} : \frac{5 π + 12 πn}{3}

\frac{5 π}{3} + \frac{12 πn}{3}

Apply rule

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

= \frac{5 π + 12 πn}{3}

x = \frac{5 π + 12 πn}{3}

x = \frac{5 π + 12 πn}{3}

x = \frac{5 π + 12 πn}{3}

- (\frac{x}{2}) = π - (\frac{π}{2} - (x - \frac{π}{3})) + 2 πn : x = - \frac{π + 12 πn}{9}

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

Move

(\frac{π}{2} - (x - \frac{π}{3}))

to the left side

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

Add

(\frac{π}{2} - (x - \frac{π}{3}))

to both sides

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

Simplify

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

Simplify

- \frac{x}{2} + \frac{π}{2} - (x - \frac{π}{3}) : \frac{- 9 x + 5 π}{6}

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

Combine the fractions

- \frac{x}{2} + \frac{π}{2} : \frac{- x + π}{2}

Apply rule

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

= \frac{- x + π}{2}

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

- (x - \frac{π}{3}) : - x + \frac{π}{3}

- (x - \frac{π}{3})

Distribute parentheses

= - (x) - (- \frac{π}{3})

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - x + \frac{π}{3}

= \frac{- x + π}{2} - x + \frac{π}{3}

Simplify

\frac{- x + π}{2} - x + \frac{π}{3} : \frac{- 9 x + 5 π}{6}

\frac{- x + π}{2} - x + \frac{π}{3}

Convert element to fraction:

x = \frac{x}{1}

= \frac{- x + π}{2} + \frac{π}{3} - \frac{x}{1}

Least Common Multiplier of

2, 3, 1 : 6

2, 3, 1

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Prime factorization of

1

Compute a number comprised of factors that appear in at least one of the following:

2, 3, 1

= 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{- x + π}{2} :

multiply the denominator and numerator by

3

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

For

\frac{x}{1} :

multiply the denominator and numerator by

6

\frac{x}{1} = \frac{x \cdot 6}{1 \cdot 6} = \frac{x \cdot 6}{6}

= \frac{( - x + π ) \cdot 3}{6} + \frac{π 2}{6} - \frac{x \cdot 6}{6}

Since the denominators are equal, combine the fractions:

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

= \frac{( - x + π ) \cdot 3 + π 2 - x \cdot 6}{6}

Expand

(- x + π) \cdot 3 + π 2 - x \cdot 6 : - 9 x + 5 π

(- x + π) \cdot 3 + π 2 - x \cdot 6

= 3 (- x + π) + 2 π - 6 x

Expand

3 (- x + π) : - 3 x + 3 π

3 (- x + π)

Apply the distributive law:

a (b + c) = ab + a c

a = 3, b = - x, c = π

= 3 (- x) + 3 π

Apply minus-plus rules

+ (- a) = - a

= - 3 x + 3 π

= - 3 x + 3 π + π 2 - x \cdot 6

Simplify

- 3 x + 3 π + π 2 - x \cdot 6 : - 9 x + 5 π

- 3 x + 3 π + π 2 - x \cdot 6

Group like terms

= - 3 x - 6 x + 3 π + 2 π

Add similar elements:

- 3 x - 6 x = - 9 x

= - 9 x + 3 π + 2 π

Add similar elements:

3 π + 2 π = 5 π

= - 9 x + 5 π

= - 9 x + 5 π

= \frac{- 9 x + 5 π}{6}

= \frac{- 9 x + 5 π}{6}

Simplify

π - (\frac{π}{2} - (x - \frac{π}{3})) + 2 πn + \frac{π}{2} - (x - \frac{π}{3}) : π + 2 πn

π - (\frac{π}{2} - (x - \frac{π}{3})) + 2 πn + \frac{π}{2} - (x - \frac{π}{3})

Add similar elements:

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

= π + 2 πn

\frac{- 9 x + 5 π}{6} = π + 2 πn

\frac{- 9 x + 5 π}{6} = π + 2 πn

\frac{- 9 x + 5 π}{6} = π + 2 πn

Multiply both sides by

6

\frac{- 9 x + 5 π}{6} = π + 2 πn

Multiply both sides by

6

\frac{6 ( - 9 x + 5 π )}{6} = 6 π + 6 \cdot 2 πn

Simplify

- 9 x + 5 π = 6 π + 12 πn

- 9 x + 5 π = 6 π + 12 πn

Move

5 π

to the right side

- 9 x + 5 π = 6 π + 12 πn

Subtract

5 π

from both sides

- 9 x + 5 π - 5 π = 6 π + 12 πn - 5 π

Simplify

- 9 x = π + 12 πn

- 9 x = π + 12 πn

Divide both sides by

- 9

- 9 x = π + 12 πn

Divide both sides by

- 9

\frac{- 9 x}{- 9} = \frac{π}{- 9} + \frac{12 πn}{- 9}

Simplify

\frac{- 9 x}{- 9} = \frac{π}{- 9} + \frac{12 πn}{- 9}

Simplify

\frac{- 9 x}{- 9} : x

\frac{- 9 x}{- 9}

Apply the fraction rule:

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

= \frac{9 x}{9}

Divide the numbers:

\frac{9}{9} = 1

= x

Simplify

\frac{π}{- 9} + \frac{12 πn}{- 9} : - \frac{π + 12 πn}{9}

\frac{π}{- 9} + \frac{12 πn}{- 9}

Apply rule

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

= \frac{π + 12 πn}{- 9}

Apply the fraction rule:

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

= - \frac{π + 12 πn}{9}

x = - \frac{π + 12 πn}{9}

x = - \frac{π + 12 πn}{9}

x = - \frac{π + 12 πn}{9}

Since the equation is undefined for:

\frac{5 π + 12 πn}{3}, - \frac{π + 12 πn}{9}

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for sin(x/2)=-cos(x-pi/3) ?
The general solution for sin(x/2)=-cos(x-pi/3) is No Solution for x\in\mathbb{R}