What is the general solution for sqrt(2)cos(x-pi/4)-1=0 ?

The general solution for sqrt(2)cos(x-pi/4)-1=0 is x= pi/2+2pin,x=2pin+2pi

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sqrt(2)cos(x-pi/4)-1=0

Solution

2 cos (x - \frac{π}{4}) - 1 = 0

Solution

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

Degrees

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

Solution steps

2 cos (x - \frac{π}{4}) - 1 = 0

Move

1

to the right side

2 cos (x - \frac{π}{4}) - 1 = 0

Add

1

to both sides

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

Simplify

2 cos (x - \frac{π}{4}) = 1

2 cos (x - \frac{π}{4}) = 1

Divide both sides by

2

2 cos (x - \frac{π}{4}) = 1

Divide both sides by

2

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

Simplify

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

Simplify

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

\frac{2 cos ( x - \frac{π}{4} )}{2}

Cancel the common factor:

2

= cos (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}

cos (x - \frac{π}{4}) = \frac{2}{2}

cos (x - \frac{π}{4}) = \frac{2}{2}

cos (x - \frac{π}{4}) = \frac{2}{2}

General solutions for

cos (x - \frac{π}{4}) = \frac{2}{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}

x - \frac{π}{4} = \frac{π}{4} + 2 πn, x - \frac{π}{4} = \frac{7 π}{4} + 2 πn

x - \frac{π}{4} = \frac{π}{4} + 2 πn, x - \frac{π}{4} = \frac{7 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{7 π}{4} + 2 πn : x = 2 πn + 2 π

x - \frac{π}{4} = \frac{7 π}{4} + 2 πn

Move

\frac{π}{4}

to the right side

x - \frac{π}{4} = \frac{7 π}{4} + 2 πn

Add

\frac{π}{4}

to both sides

x - \frac{π}{4} + \frac{π}{4} = \frac{7 π}{4} + 2 πn + \frac{π}{4}

Simplify

x - \frac{π}{4} + \frac{π}{4} = \frac{7 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{7 π}{4} + 2 πn + \frac{π}{4} : 2 πn + 2 π

\frac{7 π}{4} + 2 πn + \frac{π}{4}

Group like terms

= 2 πn + \frac{π}{4} + \frac{7 π}{4}

Combine the fractions

\frac{π}{4} + \frac{7 π}{4} : 2 π

Apply rule

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

= \frac{π + 7 π}{4}

Add similar elements:

π + 7 π = 8 π

= \frac{8 π}{4}

Divide the numbers:

\frac{8}{4} = 2

= 2 π

= 2 πn + 2 π

x = 2 πn + 2 π

x = 2 πn + 2 π

x = 2 πn + 2 π

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

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Frequently Asked Questions (FAQ)

What is the general solution for sqrt(2)cos(x-pi/4)-1=0 ?
The general solution for sqrt(2)cos(x-pi/4)-1=0 is x= pi/2+2pin,x=2pin+2pi