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

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

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sin(x+(3pi)/2)-cos(x+pi/2)=1

Solution

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

Solution

x = \frac{π}{2} + 2 πn, x = 2 πn + π

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Rewrite using trig identities

cos (x + \frac{π}{2})

Use the Angle Sum identity:

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

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

Simplify

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

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

cos (x) cos (\frac{π}{2}) = 0

cos (x) cos (\frac{π}{2})

Simplify

cos (\frac{π}{2}) : 0

cos (\frac{π}{2})

Use the following trivial identity:

cos (\frac{π}{2}) = 0

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}

= 0

= 0 \cdot cos (x)

Apply rule

0 \cdot a = 0

= 0

sin (x) sin (\frac{π}{2}) = sin (x)

sin (x) sin (\frac{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

= 1 \cdot sin (x)

Multiply:

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

= sin (x)

= 0 - sin (x)

0 - sin (x) = - sin (x)

= - sin (x)

= - sin (x)

Use the Angle Sum identity:

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

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

Simplify

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

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

sin (x) cos (\frac{3 π}{2}) = 0

sin (x) cos (\frac{3 π}{2})

cos (\frac{3 π}{2}) = 0

cos (\frac{3 π}{2})

Rewrite using trig identities:

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

cos (\frac{3 π}{2})

Write

cos (\frac{3 π}{2})

cos (π + \frac{π}{2})

= cos (π + \frac{π}{2})

Use the Angle Sum identity:

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

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

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

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{π}{2}) = 0

cos (\frac{π}{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}

= 0

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{π}{2}) = 1

sin (\frac{π}{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}

= 1

= (- 1) \cdot 0 - 0 \cdot 1

Simplify

= 0

= 0 \cdot sin (x)

Apply rule

0 \cdot a = 0

= 0

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

cos (x) sin (\frac{3 π}{2})

sin (\frac{3 π}{2}) = - 1

sin (\frac{3 π}{2})

Rewrite using trig identities:

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

sin (\frac{3 π}{2})

Write

sin (\frac{3 π}{2})

sin (π + \frac{π}{2})

= sin (π + \frac{π}{2})

Use the Angle Sum identity:

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

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

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

cos (\frac{π}{2}) = 0

cos (\frac{π}{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}

= 0

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:

sin (\frac{π}{2}) = 1

sin (\frac{π}{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}

= 1

= 0 \cdot 0 + (- 1) \cdot 1

Simplify

= - 1

= (- 1) cos (x)

Refine

= - cos (x)

= 0 - cos (x)

0 - cos (x) = - cos (x)

= - cos (x)

= - cos (x)

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

Simplify

- cos (x) - (- sin (x)) : - cos (x) + sin (x)

- cos (x) - (- sin (x))

Apply rule

- (- a) = a

= - cos (x) + sin (x)

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

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

Subtract

1

from both sides

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

Rewrite using trig identities

- cos (x) + sin (x) - 1

sin (x) - cos (x) = 2 sin (x - \frac{π}{4})

sin (x) - cos (x)

Rewrite as

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

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{1}{2}

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{1}{2}

= 2 (cos (\frac{π}{4}) sin (x) - sin (\frac{π}{4}) cos (x))

Use the Angle Sum identity:

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

= 2 sin (x - \frac{π}{4})

= - 1 + 2 sin (x - \frac{π}{4})

- 1 + 2 sin (x - \frac{π}{4}) = 0

Move

1

to the right side

- 1 + 2 sin (x - \frac{π}{4}) = 0

Add

1

to both sides

- 1 + 2 sin (x - \frac{π}{4}) + 1 = 0 + 1

Simplify

2 sin (x - \frac{π}{4}) = 1

2 sin (x - \frac{π}{4}) = 1

Divide both sides by

2

2 sin (x - \frac{π}{4}) = 1

Divide both sides by

2

\frac{2 sin ( x - \frac{π}{4} )}{2} = \frac{1}{2}

Simplify

\frac{2 sin ( x - \frac{π}{4} )}{2} = \frac{1}{2}

Simplify

\frac{2 sin ( x - \frac{π}{4} )}{2} : sin (x - \frac{π}{4})

\frac{2 sin ( x - \frac{π}{4} )}{2}

Cancel the common factor:

2

= sin (x - \frac{π}{4})

Simplify

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

sin (x - \frac{π}{4}) = \frac{2}{2}

sin (x - \frac{π}{4}) = \frac{2}{2}

sin (x - \frac{π}{4}) = \frac{2}{2}

General solutions for

sin (x - \frac{π}{4}) = \frac{2}{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}

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

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

Solve

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

x - \frac{π}{4} = \frac{π}{4} + 2 πn

Move

\frac{π}{4}

to the right side

x - \frac{π}{4} = \frac{π}{4} + 2 πn

Add

\frac{π}{4}

to both sides

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

Simplify

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

Simplify

x - \frac{π}{4} + \frac{π}{4} : x

x - \frac{π}{4} + \frac{π}{4}

Add similar elements:

- \frac{π}{4} + \frac{π}{4} = 0

= x

Simplify

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

\frac{π}{4} + 2 πn + \frac{π}{4}

Group like terms

= \frac{π}{4} + \frac{π}{4} + 2 πn

Combine the fractions

\frac{π}{4} + \frac{π}{4} : \frac{π}{2}

Apply rule

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

= \frac{π + π}{4}

Add similar elements:

π + π = 2 π

= \frac{2 π}{4}

Cancel the common factor:

2

= \frac{π}{2}

= \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

Solve

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

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

Move

\frac{π}{4}

to the right side

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

Add

\frac{π}{4}

to both sides

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

Simplify

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

Simplify

x - \frac{π}{4} + \frac{π}{4} : x

x - \frac{π}{4} + \frac{π}{4}

Add similar elements:

- \frac{π}{4} + \frac{π}{4} = 0

= x

Simplify

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

\frac{3 π}{4} + 2 πn + \frac{π}{4}

Group like terms

= 2 πn + \frac{π}{4} + \frac{3 π}{4}

Combine the fractions

\frac{π}{4} + \frac{3 π}{4} : π

Apply rule

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

= \frac{π + 3 π}{4}

Add similar elements:

π + 3 π = 4 π

= \frac{4 π}{4}

Divide the numbers:

\frac{4}{4} = 1

= π

= 2 πn + π

x = 2 πn + π

x = 2 πn + π

x = 2 πn + π

x = \frac{π}{2} + 2 πn, x = 2 πn + π

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Frequently Asked Questions (FAQ)

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