What is the general solution for 1=cos(pi/(64)*t^2) ?

The general solution for 1=cos(pi/(64)*t^2) is t=8sqrt(2)sqrt(n),t=-8sqrt(2)sqrt(n)

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1=cos(pi/(64)*t^2)

Solution

1 = cos (\frac{π}{64} \cdot t^{2})

Solution

t = 82 n, t = - 82 n

Solution steps

1 = cos (\frac{π}{64} t^{2})

Switch sides

cos (\frac{π}{64} t^{2}) = 1

General solutions for

cos (\frac{π}{64} t^{2}) = 1

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}

\frac{π}{64} t^{2} = 0 + 2 πn

\frac{π}{64} t^{2} = 0 + 2 πn

Solve

\frac{π}{64} t^{2} = 0 + 2 πn : t = 82 n, t = - 82 n

\frac{π}{64} t^{2} = 0 + 2 πn

0 + 2 πn = 2 πn

\frac{π}{64} t^{2} = 2 πn

Multiply both sides by

64

\frac{π}{64} t^{2} = 2 πn

Multiply both sides by

64

64 \cdot \frac{π}{64} t^{2} = 64 \cdot 2 πn

Simplify

π t^{2} = 128 πn

π t^{2} = 128 πn

Divide both sides by

π

π t^{2} = 128 πn

Divide both sides by

π

\frac{π t ^{2}}{π} = \frac{128 πn}{π}

Simplify

t^{2} = 128 n

t^{2} = 128 n

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

t = 128 n, t = - 128 n

Simplify

128 n : 82 n

128 n

Apply radical rule:

ab = a b,

assuming

a \geq 0, b \geq 0

128 n = 128 n

= 128 n

128 = 82

128

Prime factorization of

128 : 2^{7}

128

128

divides by

2128 = 64 \cdot 2

= 2 \cdot 64

64

divides by

264 = 32 \cdot 2

= 2 \cdot 2 \cdot 32

32

divides by

232 = 16 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 16

16

divides by

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

= 2^{7}

= 2^{7}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

2^{7} = 2^{6} \cdot 2

= 2^{6} \cdot 2

Apply radical rule:

ab = a b, a \geq 0, b \geq 0

2^{6} \cdot 2 = 2^{6} 2

= 2^{6} 2

2^{6} = 8

2^{6}

Apply radical rule:

2^{6} = 2^{\frac{6}{2}}

= 2^{\frac{6}{2}}

Divide the numbers:

\frac{6}{2} = 3

= 2^{3}

2^{3} = 8

= 8

= 82

= 82 n

Simplify

- 128 n : - 82 n

- 128 n

Apply radical rule:

ab = a b,

assuming

a \geq 0, b \geq 0

128 n = 128 n

= - 128 n

128 = 82

128

Prime factorization of

128 : 2^{7}

128

128

divides by

2128 = 64 \cdot 2

= 2 \cdot 64

64

divides by

264 = 32 \cdot 2

= 2 \cdot 2 \cdot 32

32

divides by

232 = 16 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 16

16

divides by

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

= 2^{7}

= 2^{7}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

2^{7} = 2^{6} \cdot 2

= 2^{6} \cdot 2

Apply radical rule:

ab = a b, a \geq 0, b \geq 0

2^{6} \cdot 2 = 2^{6} 2

= 2^{6} 2

2^{6} = 8

2^{6}

Apply radical rule:

2^{6} = 2^{\frac{6}{2}}

= 2^{\frac{6}{2}}

Divide the numbers:

\frac{6}{2} = 3

= 2^{3}

2^{3} = 8

= 8

= 82

= - 82 n

t = 82 n, t = - 82 n

t = 82 n, t = - 82 n

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Frequently Asked Questions (FAQ)

What is the general solution for 1=cos(pi/(64)*t^2) ?
The general solution for 1=cos(pi/(64)*t^2) is t=8sqrt(2)sqrt(n),t=-8sqrt(2)sqrt(n)