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

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

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Popular Trigonometry >

sin(x+pi)-sin(x)-1=0

Solution

sin (x + π) - sin (x) - 1 = 0

Solution

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

Degrees

x = - 3 0^{\circ} + 36 0^{\circ} n, x = 21 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (x + π) - sin (x) - 1 = 0

Rewrite using trig identities

sin (x + π) - sin (x) - 1

Use the Sum to Product identity:

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

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

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

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

x + π - x = π

x + π - x

Group like terms

= x - x + π

Add similar elements:

x - x = 0

= π

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

x + π + x = 2 x + π

x + π + x

Group like terms

= x + x + π

Add similar elements:

x + x = 2 x

= 2 x + π

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

Simplify

sin (\frac{π}{2}) : 1

sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 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}

= 1

= 2 \cdot 1 \cdot cos (\frac{2 x + π}{2})

Multiply the numbers:

2 \cdot 1 = 2

= 2 cos (\frac{2 x + π}{2})

= - 1 + 2 cos (\frac{2 x + π}{2})

- 1 + 2 cos (\frac{2 x + π}{2}) = 0

Move

1

to the right side

- 1 + 2 cos (\frac{2 x + π}{2}) = 0

Add

1

to both sides

- 1 + 2 cos (\frac{2 x + π}{2}) + 1 = 0 + 1

Simplify

2 cos (\frac{2 x + π}{2}) = 1

2 cos (\frac{2 x + π}{2}) = 1

Divide both sides by

2

2 cos (\frac{2 x + π}{2}) = 1

Divide both sides by

2

\frac{2 cos ( \frac{2 x + π}{2} )}{2} = \frac{1}{2}

Simplify

cos (\frac{2 x + π}{2}) = \frac{1}{2}

cos (\frac{2 x + π}{2}) = \frac{1}{2}

General solutions for

cos (\frac{2 x + π}{2}) = \frac{1}{2}

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

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

Solve

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

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

Multiply both sides by

2

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

Multiply both sides by

2

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

Simplify

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

Simplify

\frac{2 ( 2 x + π )}{2} : 2 x + π

\frac{2 ( 2 x + π )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 2 x + π

Simplify

2 \cdot \frac{π}{3} + 2 \cdot 2 πn : \frac{2 π}{3} + 4 πn

2 \cdot \frac{π}{3} + 2 \cdot 2 πn

Multiply

2 \cdot \frac{π}{3} : \frac{2 π}{3}

2 \cdot \frac{π}{3}

Multiply fractions:

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

= \frac{π 2}{3}

= \frac{2 π}{3} + 2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 π}{3} + 4 πn

2 x + π = \frac{2 π}{3} + 4 πn

2 x + π = \frac{2 π}{3} + 4 πn

2 x + π = \frac{2 π}{3} + 4 πn

Move

π

to the right side

2 x + π = \frac{2 π}{3} + 4 πn

Subtract

π

from both sides

2 x + π - π = \frac{2 π}{3} + 4 πn - π

Simplify

2 x = \frac{2 π}{3} + 4 πn - π

2 x = \frac{2 π}{3} + 4 πn - π

Divide both sides by

2

2 x = \frac{2 π}{3} + 4 πn - π

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{2 π}{3}}{2} + \frac{4 πn}{2} - \frac{π}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{2 π}{3}}{2} + \frac{4 πn}{2} - \frac{π}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{2 π}{3}}{2} + \frac{4 πn}{2} - \frac{π}{2} : 2 πn - \frac{π}{2} + \frac{π}{3}

\frac{\frac{2 π}{3}}{2} + \frac{4 πn}{2} - \frac{π}{2}

Group like terms

= - \frac{π}{2} + \frac{4 πn}{2} + \frac{\frac{2 π}{3}}{2}

\frac{4 πn}{2} = 2 πn

\frac{4 πn}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2 πn

\frac{\frac{2 π}{3}}{2} = \frac{π}{3}

\frac{\frac{2 π}{3}}{2}

Apply the fraction rule:

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

= \frac{2 π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{2 π}{6}

Cancel the common factor:

2

= \frac{π}{3}

= - \frac{π}{2} + 2 πn + \frac{π}{3}

Group like terms

= 2 πn - \frac{π}{2} + \frac{π}{3}

x = 2 πn - \frac{π}{2} + \frac{π}{3}

x = 2 πn - \frac{π}{2} + \frac{π}{3}

x = 2 πn - \frac{π}{2} + \frac{π}{3}

Solve

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

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

Multiply both sides by

2

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

Multiply both sides by

2

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

Simplify

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

Simplify

\frac{2 ( 2 x + π )}{2} : 2 x + π

\frac{2 ( 2 x + π )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 2 x + π

Simplify

2 \cdot \frac{5 π}{3} + 2 \cdot 2 πn : \frac{10 π}{3} + 4 πn

2 \cdot \frac{5 π}{3} + 2 \cdot 2 πn

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

2 \cdot \frac{5 π}{3}

Multiply fractions:

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

= \frac{5 π 2}{3}

Multiply the numbers:

5 \cdot 2 = 10

= \frac{10 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{10 π}{3} + 4 πn

2 x + π = \frac{10 π}{3} + 4 πn

2 x + π = \frac{10 π}{3} + 4 πn

2 x + π = \frac{10 π}{3} + 4 πn

Move

π

to the right side

2 x + π = \frac{10 π}{3} + 4 πn

Subtract

π

from both sides

2 x + π - π = \frac{10 π}{3} + 4 πn - π

Simplify

2 x = \frac{10 π}{3} + 4 πn - π

2 x = \frac{10 π}{3} + 4 πn - π

Divide both sides by

2

2 x = \frac{10 π}{3} + 4 πn - π

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{10 π}{3}}{2} + \frac{4 πn}{2} - \frac{π}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{10 π}{3}}{2} + \frac{4 πn}{2} - \frac{π}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{10 π}{3}}{2} + \frac{4 πn}{2} - \frac{π}{2} : 2 πn - \frac{π}{2} + \frac{5 π}{3}

\frac{\frac{10 π}{3}}{2} + \frac{4 πn}{2} - \frac{π}{2}

Group like terms

= - \frac{π}{2} + \frac{4 πn}{2} + \frac{\frac{10 π}{3}}{2}

\frac{4 πn}{2} = 2 πn

\frac{4 πn}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2 πn

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

\frac{\frac{10 π}{3}}{2}

Apply the fraction rule:

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

= \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}

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

Group like terms

= 2 πn - \frac{π}{2} + \frac{5 π}{3}

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

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

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

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

Graph

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Frequently Asked Questions (FAQ)

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