What is the general solution for solvefor t,0=(1.8)cos(2pift) ?

The general solution for solvefor t,0=(1.8)cos(2pift) is t= 1/(4f)+n/f ,t= 3/(4f)+n/f ;\quad f\ne 0

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solvefor t,0=(1.8)cos(2pift)

Solution

solve for

t, 0 = (1.8) cos (2 π f t)

Solution

t = \frac{1}{4 f} + \frac{n}{f}, t = \frac{3}{4 f} + \frac{n}{f}; f \neq = 0

Solution steps

0 = (1.8) cos (2 π f t)

Switch sides

1.8 cos (2 π f t) = 0

Multiply both sides by

10

1.8 cos (2 π f t) = 0

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere is one digit to the right of the decimal point, therefore multiply by

10

1.8 cos (2 π f t) \cdot 10 = 0 \cdot 10

Refine

18 cos (2 π f t) = 0

18 cos (2 π f t) = 0

Divide both sides by

18

18 cos (2 π f t) = 0

Divide both sides by

18

\frac{18 cos ( 2 π f t )}{18} = \frac{0}{18}

Simplify

cos (2 π f t) = 0

cos (2 π f t) = 0

General solutions for

cos (2 π f t) = 0

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}

2 π f t = \frac{π}{2} + 2 πn, 2 π f t = \frac{3 π}{2} + 2 πn

2 π f t = \frac{π}{2} + 2 πn, 2 π f t = \frac{3 π}{2} + 2 πn

Solve

2 π f t = \frac{π}{2} + 2 πn : t = \frac{1}{4 f} + \frac{n}{f}; f \neq = 0

2 π f t = \frac{π}{2} + 2 πn

Divide both sides by

2 π f; f \neq = 0

2 π f t = \frac{π}{2} + 2 πn

Divide both sides by

2 π f; f \neq = 0

\frac{2 π f t}{2 π f} = \frac{\frac{π}{2}}{2 π f} + \frac{2 πn}{2 π f}; f \neq = 0

Simplify

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

Simplify

\frac{2 π f t}{2 π f} : t

\frac{2 π f t}{2 π f}

Divide the numbers:

\frac{2}{2} = 1

= \frac{π t f}{π f}

Cancel the common factor:

π

= \frac{t f}{f}

Cancel the common factor:

f

= t

Simplify

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

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

\frac{\frac{π}{2}}{2 π f} = \frac{1}{4 f}

\frac{\frac{π}{2}}{2 π f}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 2 π f}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4 π f}

Cancel the common factor:

π

= \frac{1}{4 f}

\frac{2 πn}{2 π f} = \frac{n}{f}

\frac{2 πn}{2 π f}

Cancel

\frac{2 πn}{2 π f} : \frac{n}{f}

\frac{2 πn}{2 π f}

Divide the numbers:

\frac{2}{2} = 1

= \frac{πn}{π f}

Cancel the common factor:

π

= \frac{n}{f}

= \frac{n}{f}

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

t = \frac{1}{4 f} + \frac{n}{f}; f \neq = 0

t = \frac{1}{4 f} + \frac{n}{f}; f \neq = 0

t = \frac{1}{4 f} + \frac{n}{f}; f \neq = 0

Solve

2 π f t = \frac{3 π}{2} + 2 πn : t = \frac{3}{4 f} + \frac{n}{f}; f \neq = 0

2 π f t = \frac{3 π}{2} + 2 πn

Divide both sides by

2 π f; f \neq = 0

2 π f t = \frac{3 π}{2} + 2 πn

Divide both sides by

2 π f; f \neq = 0

\frac{2 π f t}{2 π f} = \frac{\frac{3 π}{2}}{2 π f} + \frac{2 πn}{2 π f}; f \neq = 0

Simplify

\frac{2 π f t}{2 π f} = \frac{\frac{3 π}{2}}{2 π f} + \frac{2 πn}{2 π f}

Simplify

\frac{2 π f t}{2 π f} : t

\frac{2 π f t}{2 π f}

Divide the numbers:

\frac{2}{2} = 1

= \frac{π t f}{π f}

Cancel the common factor:

π

= \frac{t f}{f}

Cancel the common factor:

f

= t

Simplify

\frac{\frac{3 π}{2}}{2 π f} + \frac{2 πn}{2 π f} : \frac{3}{4 f} + \frac{n}{f}

\frac{\frac{3 π}{2}}{2 π f} + \frac{2 πn}{2 π f}

\frac{\frac{3 π}{2}}{2 π f} = \frac{3}{4 f}

\frac{\frac{3 π}{2}}{2 π f}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 2 π f}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 π}{4 π f}

Cancel the common factor:

π

= \frac{3}{4 f}

\frac{2 πn}{2 π f} = \frac{n}{f}

\frac{2 πn}{2 π f}

Cancel

\frac{2 πn}{2 π f} : \frac{n}{f}

\frac{2 πn}{2 π f}

Divide the numbers:

\frac{2}{2} = 1

= \frac{πn}{π f}

Cancel the common factor:

π

= \frac{n}{f}

= \frac{n}{f}

= \frac{3}{4 f} + \frac{n}{f}

t = \frac{3}{4 f} + \frac{n}{f}; f \neq = 0

t = \frac{3}{4 f} + \frac{n}{f}; f \neq = 0

t = \frac{3}{4 f} + \frac{n}{f}; f \neq = 0

t = \frac{1}{4 f} + \frac{n}{f}, t = \frac{3}{4 f} + \frac{n}{f}; f \neq = 0

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Frequently Asked Questions (FAQ)

What is the general solution for solvefor t,0=(1.8)cos(2pift) ?
The general solution for solvefor t,0=(1.8)cos(2pift) is t= 1/(4f)+n/f ,t= 3/(4f)+n/f ;\quad f\ne 0