What is the general solution for sin(2pit)=1 ?

The general solution for sin(2pit)=1 is t= 1/4+n

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sin(2pit)=1

sin (2 π t) = 1

t = \frac{1}{4} + n

Degrees

t = 14.32394 \dots^{\circ}

Solution steps

sin (2 π t) = 1

General solutions for

sin (2 π t) = 1

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 π t = \frac{π}{2} + 2 πn

2 π t = \frac{π}{2} + 2 πn

Solve

2 π t = \frac{π}{2} + 2 πn : t = \frac{1}{4} + n

2 π t = \frac{π}{2} + 2 πn

Divide both sides by

2 π

2 π t = \frac{π}{2} + 2 πn

Divide both sides by

2 π

\frac{2 π t}{2 π} = \frac{\frac{π}{2}}{2 π} + \frac{2 πn}{2 π}

Simplify

\frac{2 π t}{2 π} = \frac{\frac{π}{2}}{2 π} + \frac{2 πn}{2 π}

Simplify

\frac{2 π t}{2 π} : t

\frac{2 π t}{2 π}

Divide the numbers:

\frac{2}{2} = 1

= \frac{π t}{π}

Cancel the common factor:

π

= t

Simplify

\frac{\frac{π}{2}}{2 π} + \frac{2 πn}{2 π} : \frac{1}{4} + n

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

\frac{\frac{π}{2}}{2 π} = \frac{1}{4}

\frac{\frac{π}{2}}{2 π}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 2 π}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4 π}

Cancel the common factor:

π

= \frac{1}{4}

\frac{2 πn}{2 π} = n

\frac{2 πn}{2 π}

Cancel

\frac{2 πn}{2 π} : n

\frac{2 πn}{2 π}

Divide the numbers:

\frac{2}{2} = 1

= \frac{πn}{π}

Cancel the common factor:

π

= n

= n

= \frac{1}{4} + n

t = \frac{1}{4} + n

t = \frac{1}{4} + n

t = \frac{1}{4} + n

t = \frac{1}{4} + n

What is the general solution for sin(2pit)=1 ?
The general solution for sin(2pit)=1 is t= 1/4+n