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+(2pi)/3 ,x=2pin-pi/2+(4pi)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

sin (x + π) - sin (x) + 1 = 0

Solution

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

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 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

Subtract

1

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

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

Solve

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

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

Multiply both sides by

2

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

Multiply both sides by

2

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

Simplify

\frac{2 ( 2 x + π )}{2} = 2 \cdot \frac{2 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{2 π}{3} + 2 \cdot 2 πn : \frac{4 π}{3} + 4 πn

2 \cdot \frac{2 π}{3} + 2 \cdot 2 πn

2 \cdot \frac{2 π}{3} = \frac{4 π}{3}

2 \cdot \frac{2 π}{3}

Multiply fractions:

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

= \frac{2 π 2}{3}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{4 π}{3} + 4 πn

2 x + π = \frac{4 π}{3} + 4 πn

2 x + π = \frac{4 π}{3} + 4 πn

2 x + π = \frac{4 π}{3} + 4 πn

Move

π

to the right side

2 x + π = \frac{4 π}{3} + 4 πn

Subtract

π

from both sides

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

Simplify

2 x = \frac{4 π}{3} + 4 πn - π

2 x = \frac{4 π}{3} + 4 πn - π

Divide both sides by

2

2 x = \frac{4 π}{3} + 4 πn - π

Divide both sides by

2

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

Simplify

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

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

Group like terms

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

\frac{4 πn}{2} = 2 πn

\frac{4 πn}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2 πn

\frac{\frac{4 π}{3}}{2} = \frac{2 π}{3}

\frac{\frac{4 π}{3}}{2}

Apply the fraction rule:

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

= \frac{4 π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{4 π}{6}

Cancel the common factor:

2

= \frac{2 π}{3}

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

Group like terms

= 2 πn - \frac{π}{2} + \frac{2 π}{3}

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

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

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

Solve

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

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

Multiply both sides by

2

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

Multiply both sides by

2

\frac{2 ( 2 x + π )}{2} = 2 \cdot \frac{4 π}{3} + 2 \cdot 2 πn

Simplify

\frac{2 ( 2 x + π )}{2} = 2 \cdot \frac{4 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{4 π}{3} + 2 \cdot 2 πn : \frac{8 π}{3} + 4 πn

2 \cdot \frac{4 π}{3} + 2 \cdot 2 πn

2 \cdot \frac{4 π}{3} = \frac{8 π}{3}

2 \cdot \frac{4 π}{3}

Multiply fractions:

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

= \frac{4 π 2}{3}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{8 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{8 π}{3} + 4 πn

2 x + π = \frac{8 π}{3} + 4 πn

2 x + π = \frac{8 π}{3} + 4 πn

2 x + π = \frac{8 π}{3} + 4 πn

Move

π

to the right side

2 x + π = \frac{8 π}{3} + 4 πn

Subtract

π

from both sides

2 x + π - π = \frac{8 π}{3} + 4 πn - π

Simplify

2 x = \frac{8 π}{3} + 4 πn - π

2 x = \frac{8 π}{3} + 4 πn - π

Divide both sides by

2

2 x = \frac{8 π}{3} + 4 πn - π

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{8 π}{3}}{2} + \frac{4 πn}{2} - \frac{π}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{8 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{8 π}{3}}{2} + \frac{4 πn}{2} - \frac{π}{2} : 2 πn - \frac{π}{2} + \frac{4 π}{3}

\frac{\frac{8 π}{3}}{2} + \frac{4 πn}{2} - \frac{π}{2}

Group like terms

= - \frac{π}{2} + \frac{4 πn}{2} + \frac{\frac{8 π}{3}}{2}

\frac{4 πn}{2} = 2 πn

\frac{4 πn}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2 πn

\frac{\frac{8 π}{3}}{2} = \frac{4 π}{3}

\frac{\frac{8 π}{3}}{2}

Apply the fraction rule:

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

= \frac{8 π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{8 π}{6}

Cancel the common factor:

2

= \frac{4 π}{3}

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

Group like terms

= 2 πn - \frac{π}{2} + \frac{4 π}{3}

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

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

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

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

Popular Examples

sin(x)=sqrt(3)-sin(x)

sin (x) = 3 - sin (x)

cot(θ)=-(sqrt(3))/3

cot (θ) = - \frac{3}{3}

sin(x)-tan(x)=0

sin (x) - tan (x) = 0

sin(x+20)=cos(x)

sin (x + 2 0^{\circ}) = cos (x)

cos(x)-sqrt(3)sin(x)=1

cos (x) - 3 sin (x) = 1

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+(2pi)/3 ,x=2pin-pi/2+(4pi)/3