What is the general solution for sin(1/x)=0,5 ?

The general solution for sin(1/x)=0,5 is No Solution for x\in\mathbb{R};\quad n\ne 0,n\ne-1/2

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sin(1/x)=0,5

Solution

sin (\frac{1}{x}) = 0, 5

Solution

No Solution for x \in R; n \neq = 0, n \neq = - \frac{1}{2}

Solution steps

sin (\frac{1}{x}) = 0, 5

General solutions for

sin (\frac{1}{x}) = 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{1}{x} = 0 + 2 πn, \frac{1}{x} = π + 2 πn

\frac{1}{x} = 0 + 2 πn, \frac{1}{x} = π + 2 πn

Solve

\frac{1}{x} = 0 + 2 πn : x = \frac{1}{2 πn}; n \neq = 0

\frac{1}{x} = 0 + 2 πn

Multiply both sides by

x

\frac{1}{x} = 0 + 2 πn

Multiply both sides by

x

\frac{1}{x} x = 0 \cdot x + 2 πn x

Simplify

1 = 0 + 2 πn x

Simplify

0 + 2 πn x : 2 πn x

0 + 2 πn x

0 + 2 πn x = 2 πn x

= 2 πn x

1 = 2 πn x

1 = 2 πn x

Switch sides

2 πn x = 1

Divide both sides by

2 πn; n \neq = 0

2 πn x = 1

Divide both sides by

2 πn; n \neq = 0

\frac{2 πn x}{2 πn} = \frac{1}{2 πn}; n \neq = 0

Simplify

x = \frac{1}{2 πn}; n \neq = 0

x = \frac{1}{2 πn}; n \neq = 0

Solve

\frac{1}{x} = π + 2 πn : x = \frac{1}{π ( 1 + 2 n )}; n \neq = - \frac{1}{2}

\frac{1}{x} = π + 2 πn

Multiply both sides by

x

\frac{1}{x} = π + 2 πn

Multiply both sides by

x

\frac{1}{x} x = π x + 2 πn x

Simplify

1 = π x + 2 πn x

1 = π x + 2 πn x

Switch sides

π x + 2 πn x = 1

Factor

π x + 2 πn x : π x (1 + 2 n)

π x + 2 πn x

Factor out common term

x π

= x π (1 + 2 n)

π x (1 + 2 n) = 1

Divide both sides by

π (1 + 2 n); n \neq = - \frac{1}{2}

π x (1 + 2 n) = 1

Divide both sides by

π (1 + 2 n); n \neq = - \frac{1}{2}

\frac{π x ( 1 + 2 n )}{π ( 1 + 2 n )} = \frac{1}{π ( 1 + 2 n )}; n \neq = - \frac{1}{2}

Simplify

x = \frac{1}{π ( 1 + 2 n )}; n \neq = - \frac{1}{2}

x = \frac{1}{π ( 1 + 2 n )}; n \neq = - \frac{1}{2}

x = \frac{1}{2 πn}, x = \frac{1}{π ( 1 + 2 n )}; n \neq = 0, n \neq = - \frac{1}{2}

Solutions for the range

5

No Solution for x \in R; n \neq = 0, n \neq = - \frac{1}{2}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for sin(1/x)=0,5 ?
The general solution for sin(1/x)=0,5 is No Solution for x\in\mathbb{R};\quad n\ne 0,n\ne-1/2