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

The general solution for sin(2x-pi/4)=0 is x=pin+pi/8 ,x=pin+pi/2+pi/8

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sin(2x-pi/4)=0

Solution

sin (2 x - \frac{π}{4}) = 0

Solution

x = πn + \frac{π}{8}, x = πn + \frac{π}{2} + \frac{π}{8}

Degrees

x = 22. 5^{\circ} + 18 0^{\circ} n, x = 112. 5^{\circ} + 18 0^{\circ} n

Solution steps

sin (2 x - \frac{π}{4}) = 0

General solutions for

sin (2 x - \frac{π}{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}

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

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

Solve

2 x - \frac{π}{4} = 0 + 2 πn : x = πn + \frac{π}{8}

2 x - \frac{π}{4} = 0 + 2 πn

0 + 2 πn = 2 πn

2 x - \frac{π}{4} = 2 πn

Move

\frac{π}{4}

to the right side

2 x - \frac{π}{4} = 2 πn

Add

\frac{π}{4}

to both sides

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

Simplify

2 x = 2 πn + \frac{π}{4}

2 x = 2 πn + \frac{π}{4}

Divide both sides by

2

2 x = 2 πn + \frac{π}{4}

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

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

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

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{π}{8}

= πn + \frac{π}{8}

x = πn + \frac{π}{8}

x = πn + \frac{π}{8}

x = πn + \frac{π}{8}

Solve

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

2 x - \frac{π}{4} = π + 2 πn

Move

\frac{π}{4}

to the right side

2 x - \frac{π}{4} = π + 2 πn

Add

\frac{π}{4}

to both sides

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

Simplify

2 x = π + 2 πn + \frac{π}{4}

2 x = π + 2 πn + \frac{π}{4}

Divide both sides by

2

2 x = π + 2 πn + \frac{π}{4}

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

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

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

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{π}{8}

= \frac{π}{2} + πn + \frac{π}{8}

Group like terms

= πn + \frac{π}{2} + \frac{π}{8}

x = πn + \frac{π}{2} + \frac{π}{8}

x = πn + \frac{π}{2} + \frac{π}{8}

x = πn + \frac{π}{2} + \frac{π}{8}

x = πn + \frac{π}{8}, x = πn + \frac{π}{2} + \frac{π}{8}

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Frequently Asked Questions (FAQ)

What is the general solution for sin(2x-pi/4)=0 ?
The general solution for sin(2x-pi/4)=0 is x=pin+pi/8 ,x=pin+pi/2+pi/8