What is the general solution for sin(x+pi/4)=-(sqrt(2))/2 ,0<= x<= 2pi ?

Q: What is the general solution for sin(x+pi/4)=-(sqrt(2))/2 ,0<= x<= 2pi ?

The general solution for sin(x+pi/4)=-(sqrt(2))/2 ,0<= x<= 2pi is x=pi,x=(3pi)/2

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sin(x+pi/4)=-(sqrt(2))/2 ,0<= x<= 2pi

Solution

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

Solution

x = π, x = \frac{3 π}{2}

Degrees

x = 18 0^{\circ}, x = 27 0^{\circ}

Solution steps

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

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

x + \frac{π}{4} = \frac{5 π}{4} + 2 πn, x + \frac{π}{4} = \frac{7 π}{4} + 2 πn

Solve

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

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

Move

\frac{π}{4}

to the right side

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

Subtract

\frac{π}{4}

from both sides

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

Simplify

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

\frac{5 π}{4} + 2 πn - \frac{π}{4}

Group like terms

= 2 πn - \frac{π}{4} + \frac{5 π}{4}

Combine the fractions

- \frac{π}{4} + \frac{5 π}{4} : π

Apply rule

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

= \frac{- π + 5 π}{4}

Add similar elements:

- π + 5 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

Solve

x + \frac{π}{4} = \frac{7 π}{4} + 2 πn : x = 2 πn + \frac{3 π}{2}

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

Subtract

\frac{π}{4}

from 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 + \frac{3 π}{2}

\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} : \frac{3 π}{2}

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 = 6 π

= \frac{6 π}{4}

Cancel the common factor:

2

= \frac{3 π}{2}

= 2 πn + \frac{3 π}{2}

x = 2 πn + \frac{3 π}{2}

x = 2 πn + \frac{3 π}{2}

x = 2 πn + \frac{3 π}{2}

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

Solutions for the range

0 \leq x \leq 2 π

x = π, x = \frac{3 π}{2}

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Frequently Asked Questions (FAQ)

What is the general solution for sin(x+pi/4)=-(sqrt(2))/2 ,0<= x<= 2pi ?
The general solution for sin(x+pi/4)=-(sqrt(2))/2 ,0<= x<= 2pi is x=pi,x=(3pi)/2