What is the general solution for cos(x+1)=sin(x) ?

The general solution for cos(x+1)=sin(x) is x=2pin+pi/4-1/2 ,x=-1/2+(5pi)/4+2pin

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cos(x+1)=sin(x)

Solution

cos (x + 1) = sin (x)

Solution

x = 2 πn + \frac{π}{4} - \frac{1}{2}, x = - \frac{1}{2} + \frac{5 π}{4} + 2 πn

Degrees

x = 16.35211 \dots^{\circ} + 36 0^{\circ} n, x = 196.35211 \dots^{\circ} + 36 0^{\circ} n

Solution steps

cos (x + 1) = sin (x)

Subtract

sin (x)

from both sides

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

Rewrite using trig identities

cos (1 + x) - sin (x)

Use the following identity:

sin (x) = cos (\frac{π}{2} - x)

= cos (1 + x) - cos (\frac{π}{2} - x)

Use the Sum to Product identity:

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

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

Simplify

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

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

\frac{1 + x + \frac{π}{2} - x}{2} = \frac{π + 2}{4}

\frac{1 + x + \frac{π}{2} - x}{2}

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

1 + x + \frac{π}{2} - x

Group like terms

= x - x + \frac{π}{2} + 1

Add similar elements:

x - x = 0

= \frac{π}{2} + 1

= \frac{\frac{π}{2} + 1}{2}

Join

\frac{π}{2} + 1 : \frac{π + 2}{2}

\frac{π}{2} + 1

Convert element to fraction:

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

= \frac{π}{2} + \frac{1 \cdot 2}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{π + 1 \cdot 2}{2}

Multiply the numbers:

1 \cdot 2 = 2

= \frac{π + 2}{2}

= \frac{\frac{π + 2}{2}}{2}

Apply the fraction rule:

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

= \frac{π + 2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π + 2}{4}

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

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

\frac{1 + x - ( \frac{π}{2} - x )}{2}

Expand

1 + x - (\frac{π}{2} - x) : 2 x - \frac{π}{2} + 1

1 + x - (\frac{π}{2} - x)

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

- (\frac{π}{2} - x)

Distribute parentheses

= - (\frac{π}{2}) - (- x)

Apply minus-plus rules

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

= - \frac{π}{2} + x

= 1 + x - \frac{π}{2} + x

Simplify

1 + x - \frac{π}{2} + x : 2 x - \frac{π}{2} + 1

1 + x - \frac{π}{2} + x

Group like terms

= x + x - \frac{π}{2} + 1

Add similar elements:

x + x = 2 x

= 2 x - \frac{π}{2} + 1

= 2 x - \frac{π}{2} + 1

= \frac{2 x - \frac{π}{2} + 1}{2}

Join

2 x - \frac{π}{2} + 1 : \frac{4 x - π + 2}{2}

2 x - \frac{π}{2} + 1

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

2 x \cdot 2 - π + 1 \cdot 2 = 4 x - π + 2

2 x \cdot 2 - π + 1 \cdot 2

Multiply the numbers:

2 \cdot 2 = 4

= 4 x - π + 1 \cdot 2

Multiply the numbers:

1 \cdot 2 = 2

= 4 x - π + 2

= \frac{4 x - π + 2}{2}

= \frac{\frac{4 x - π + 2}{2}}{2}

Apply the fraction rule:

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

= \frac{4 x - π + 2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4 x - π + 2}{4}

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

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

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

Divide both sides by

- 2 sin (\frac{π + 2}{4})

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

Divide both sides by

- 2 sin (\frac{π + 2}{4})

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

Simplify

sin (\frac{4 x - π + 2}{4}) = 0

sin (\frac{4 x - π + 2}{4}) = 0

General solutions for

sin (\frac{4 x - π + 2}{4}) = 0

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{4 x - π + 2}{4} = 0 + 2 πn, \frac{4 x - π + 2}{4} = π + 2 πn

\frac{4 x - π + 2}{4} = 0 + 2 πn, \frac{4 x - π + 2}{4} = π + 2 πn

Solve

\frac{4 x - π + 2}{4} = 0 + 2 πn : x = 2 πn + \frac{π}{4} - \frac{1}{2}

\frac{4 x - π + 2}{4} = 0 + 2 πn

0 + 2 πn = 2 πn

\frac{4 x - π + 2}{4} = 2 πn

Multiply both sides by

4

\frac{4 x - π + 2}{4} = 2 πn

Multiply both sides by

4

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

Simplify

4 x - π + 2 = 8 πn

4 x - π + 2 = 8 πn

Move

π

to the right side

4 x - π + 2 = 8 πn

Add

π

to both sides

4 x - π + 2 + π = 8 πn + π

Simplify

4 x + 2 = 8 πn + π

4 x + 2 = 8 πn + π

Move

2

to the right side

4 x + 2 = 8 πn + π

Subtract

2

from both sides

4 x + 2 - 2 = 8 πn + π - 2

Simplify

4 x = 8 πn + π - 2

4 x = 8 πn + π - 2

Divide both sides by

4

4 x = 8 πn + π - 2

Divide both sides by

4

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

Simplify

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

Simplify

\frac{4 x}{4} : x

\frac{4 x}{4}

Divide the numbers:

\frac{4}{4} = 1

= x

Simplify

\frac{8 πn}{4} + \frac{π}{4} - \frac{2}{4} : 2 πn + \frac{π}{4} - \frac{1}{2}

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

Divide the numbers:

\frac{8}{4} = 2

= 2 πn + \frac{π}{4} - \frac{2}{4}

Cancel

\frac{2}{4} : \frac{1}{2}

\frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

= 2 πn + \frac{π}{4} - \frac{1}{2}

x = 2 πn + \frac{π}{4} - \frac{1}{2}

x = 2 πn + \frac{π}{4} - \frac{1}{2}

x = 2 πn + \frac{π}{4} - \frac{1}{2}

Solve

\frac{4 x - π + 2}{4} = π + 2 πn : x = - \frac{1}{2} + \frac{5 π}{4} + 2 πn

\frac{4 x - π + 2}{4} = π + 2 πn

Multiply both sides by

4

\frac{4 x - π + 2}{4} = π + 2 πn

Multiply both sides by

4

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

Simplify

4 x - π + 2 = 4 π + 8 πn

4 x - π + 2 = 4 π + 8 πn

Move

π

to the right side

4 x - π + 2 = 4 π + 8 πn

Add

π

to both sides

4 x - π + 2 + π = 4 π + 8 πn + π

Simplify

4 x + 2 = 5 π + 8 πn

4 x + 2 = 5 π + 8 πn

Move

2

to the right side

4 x + 2 = 5 π + 8 πn

Subtract

2

from both sides

4 x + 2 - 2 = 5 π + 8 πn - 2

Simplify

4 x = 5 π + 8 πn - 2

4 x = 5 π + 8 πn - 2

Divide both sides by

4

4 x = 5 π + 8 πn - 2

Divide both sides by

4

\frac{4 x}{4} = \frac{5 π}{4} + \frac{8 πn}{4} - \frac{2}{4}

Simplify

\frac{4 x}{4} = \frac{5 π}{4} + \frac{8 πn}{4} - \frac{2}{4}

Simplify

\frac{4 x}{4} : x

\frac{4 x}{4}

Divide the numbers:

\frac{4}{4} = 1

= x

Simplify

\frac{5 π}{4} + \frac{8 πn}{4} - \frac{2}{4} : - \frac{1}{2} + \frac{5 π}{4} + 2 πn

\frac{5 π}{4} + \frac{8 πn}{4} - \frac{2}{4}

Group like terms

= - \frac{2}{4} + \frac{5 π}{4} + \frac{8 πn}{4}

Cancel

\frac{2}{4} : \frac{1}{2}

\frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

= - \frac{1}{2} + \frac{5 π}{4} + \frac{8 πn}{4}

Divide the numbers:

\frac{8}{4} = 2

= - \frac{1}{2} + \frac{5 π}{4} + 2 πn

x = - \frac{1}{2} + \frac{5 π}{4} + 2 πn

x = - \frac{1}{2} + \frac{5 π}{4} + 2 πn

x = - \frac{1}{2} + \frac{5 π}{4} + 2 πn

x = 2 πn + \frac{π}{4} - \frac{1}{2}, x = - \frac{1}{2} + \frac{5 π}{4} + 2 πn

Popular Examples

2sin^2(θ)=2+cos(2θ)

2 sin^{2} (θ) = 2 + cos (2 θ)

tan^2(x)-1=0,0<= x<= 2pi

tan^{2} (x) - 1 = 0, 0 \leq x \leq 2 π

sin^2(θ)= 1/4

sin^{2} (θ) = \frac{1}{4}

(tan(θ)+1)(sec(θ)+1)=0

(tan (θ) + 1) (sec (θ) + 1) = 0

2sin(θ)-tan^2(θ)=0

2 sin (θ) - tan^{2} (θ) = 0

Frequently Asked Questions (FAQ)

What is the general solution for cos(x+1)=sin(x) ?
The general solution for cos(x+1)=sin(x) is x=2pin+pi/4-1/2 ,x=-1/2+(5pi)/4+2pin