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

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

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

Solution

sin (x - \frac{5 π}{3}) + sin (x + \frac{5 π}{3}) = 1

Solution

x = \frac{π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (x - \frac{5 π}{3}) + sin (x + \frac{5 π}{3}) = 1

Rewrite using trig identities

sin (x - \frac{5 π}{3}) + sin (x + \frac{5 π}{3})

Use the Sum to Product identity:

sin (s) + sin (t) = 2 sin (\frac{s + t}{2}) cos (\frac{s - t}{2})

= 2 sin (\frac{x - \frac{5 π}{3} + x + \frac{5 π}{3}}{2}) cos (\frac{x - \frac{5 π}{3} - ( x + \frac{5 π}{3} )}{2})

Simplify

2 sin (\frac{x - \frac{5 π}{3} + x + \frac{5 π}{3}}{2}) cos (\frac{x - \frac{5 π}{3} - ( x + \frac{5 π}{3} )}{2}) : sin (x)

2 sin (\frac{x - \frac{5 π}{3} + x + \frac{5 π}{3}}{2}) cos (\frac{x - \frac{5 π}{3} - ( x + \frac{5 π}{3} )}{2})

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

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

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

x - \frac{5 π}{3} + x + \frac{5 π}{3}

Group like terms

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

Add similar elements:

x + x = 2 x

= 2 x - \frac{5 π}{3} + \frac{5 π}{3}

Add similar elements:

- \frac{5 π}{3} + \frac{5 π}{3} = 0

= 2 x

= \frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

= 2 sin (x) cos (\frac{x - ( x + \frac{5 π}{3} ) - \frac{5 π}{3}}{2})

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

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

Join

x - \frac{5 π}{3} - (x + \frac{5 π}{3}) : - \frac{10 π}{3}

x - \frac{5 π}{3} - (x + \frac{5 π}{3})

Convert element to fraction:

x = \frac{x 3}{3}, (x + \frac{5 π}{3}) = \frac{( x + \frac{5 π}{3} ) 3}{3}

= \frac{x \cdot 3}{3} - \frac{5 π}{3} - \frac{( x + \frac{5 π}{3} ) \cdot 3}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{x \cdot 3 - 5 π - ( x + \frac{5 π}{3} ) \cdot 3}{3}

Expand

x \cdot 3 - 5 π - (x + \frac{5 π}{3}) \cdot 3 : - 10 π

x \cdot 3 - 5 π - (x + \frac{5 π}{3}) \cdot 3

= 3 x - 5 π - 3 (x + \frac{5 π}{3})

Expand

- 3 (x + \frac{5 π}{3}) : - 3 x - 5 π

- 3 (x + \frac{5 π}{3})

Apply the distributive law:

a (b + c) = ab + a c

a = - 3, b = x, c = \frac{5 π}{3}

= - 3 x + (- 3) \frac{5 π}{3}

Apply minus-plus rules

+ (- a) = - a

= - 3 x - 3 \cdot \frac{5 π}{3}

3 \cdot \frac{5 π}{3} = 5 π

3 \cdot \frac{5 π}{3}

Multiply fractions:

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

= \frac{5 π 3}{3}

Cancel the common factor:

3

= 5 π

= - 3 x - 5 π

= x \cdot 3 - 5 π - 3 x - 5 π

Simplify

x \cdot 3 - 5 π - 3 x - 5 π : - 10 π

x \cdot 3 - 5 π - 3 x - 5 π

Group like terms

= 3 x - 3 x - 5 π - 5 π

Add similar elements:

3 x - 3 x = 0

= - 5 π - 5 π

Add similar elements:

- 5 π - 5 π = - 10 π

= - 10 π

= - 10 π

= \frac{- 10 π}{3}

Apply the fraction rule:

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

= - \frac{10 π}{3}

= \frac{- \frac{10 π}{3}}{2}

Apply the fraction rule:

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

= - \frac{\frac{10 π}{3}}{2}

Apply the fraction rule:

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

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

= - \frac{10 π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= - \frac{10 π}{6}

Cancel the common factor:

2

= - \frac{5 π}{3}

= 2 cos (- \frac{5 π}{3}) sin (x)

cos (- \frac{5 π}{3}) = \frac{1}{2}

cos (- \frac{5 π}{3})

Use the following property:

cos (- x) = cos (x)

cos (- \frac{5 π}{3}) = cos (\frac{5 π}{3})

= cos (\frac{5 π}{3})

Rewrite using trig identities:

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

cos (\frac{5 π}{3})

Write

cos (\frac{5 π}{3})

cos (π + \frac{2 π}{3})

= cos (π + \frac{2 π}{3})

Use the Angle Sum identity:

cos (s + t) = cos (s) cos (t) - sin (s) sin (t)

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

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

Use the following trivial identity:

cos (π) = (- 1)

cos (π)

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= (- 1)

Use the following trivial identity:

cos (\frac{2 π}{3}) = - \frac{1}{2}

cos (\frac{2 π}{3})

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - \frac{1}{2}

Use the following trivial identity:

sin (π) = 0

sin (π)

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}

= 0

Use the following trivial identity:

sin (\frac{2 π}{3}) = \frac{3}{2}

sin (\frac{2 π}{3})

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{3}{2}

= (- 1) (- \frac{1}{2}) - 0 \cdot \frac{3}{2}

Simplify

= \frac{1}{2}

= 2 \cdot \frac{1}{2} sin (x)

Multiply fractions:

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

= \frac{1 \cdot 2}{2} sin (x)

Cancel the common factor:

2

= sin (x) \cdot 1

Multiply:

sin (x) \cdot 1 = sin (x)

= sin (x)

= sin (x)

sin (x) = 1

General solutions for

sin (x) = 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}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

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