What is the general solution for 15=60sin(pi/(15[0-d]))+75 ?

The general solution for 15=60sin(pi/(15[0-d]))+75 is d=-2/(15(4n+3));\quad n\ne-3/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

15=60sin(pi/(15[0-d]))+75

Solution

15 = 60 sin (\frac{π}{15 [ 0 - d ]}) + 75

Solution

d = - \frac{2}{15 ( 4 n + 3 )}; n \neq = - \frac{3}{4}

Degrees

d = 0^{\circ} - 1.09134 \dots^{\circ} n

Solution steps

15 = 60 sin (\frac{π}{15 [ 0 - d ]}) + 75

Switch sides

60 sin (\frac{π}{15 [ 0 - d ]}) + 75 = 15

Move

75

to the right side

60 sin (\frac{π}{15 [ 0 - d ]}) + 75 = 15

Subtract

75

from both sides

60 sin (\frac{π}{15 [ 0 - d ]}) + 75 - 75 = 15 - 75

Simplify

60 sin (\frac{π}{15 [ 0 - d ]}) = - 60

60 sin (\frac{π}{15 [ 0 - d ]}) = - 60

Divide both sides by

60

60 sin (\frac{π}{15 [ 0 - d ]}) = - 60

Divide both sides by

60

\frac{60 sin ( \frac{π}{15 [ 0 - d ]} )}{60} = \frac{- 60}{60}

Simplify

sin (\frac{π}{15 [ 0 - d ]}) = - 1

sin (\frac{π}{15 [ 0 - d ]}) = - 1

General solutions for

sin (\frac{π}{15 [ 0 - d ]}) = - 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}

\frac{π}{15 [ 0 - d ]} = \frac{3 π}{2} + 2 πn

\frac{π}{15 [ 0 - d ]} = \frac{3 π}{2} + 2 πn

Solve

\frac{π}{15 [ 0 - d ]} = \frac{3 π}{2} + 2 πn : d = - \frac{2}{15 ( 4 n + 3 )}; n \neq = - \frac{3}{4}

\frac{π}{15 [ 0 - d ]} = \frac{3 π}{2} + 2 πn

Simplify

\frac{π}{15 ( 0 - d )} : - \frac{π}{15 d}

\frac{π}{15 [ 0 - d ]}

15 (0 - d) = - 15 d

15 (0 - d)

0 - d = - d

= 15 (- d)

Remove parentheses:

(- a) = - a

= - 15 d

= \frac{π}{- 15 d}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{π}{15 d}

- \frac{π}{15 d} = \frac{3 π}{2} + 2 πn

Multiply both sides by

d

- \frac{π}{15 d} = \frac{3 π}{2} + 2 πn

Multiply both sides by

d

- \frac{π}{15 d} d = \frac{3 π}{2} d + 2 πn d

Simplify

- \frac{π}{15} = \frac{3 π}{2} d + 2 πn d

- \frac{π}{15} = \frac{3 π}{2} d + 2 πn d

Switch sides

\frac{3 π}{2} d + 2 πn d = - \frac{π}{15}

Multiply both sides by

2

\frac{3 π}{2} d + 2 πn d = - \frac{π}{15}

Multiply both sides by

2

\frac{3 π}{2} d \cdot 2 + 2 πn d \cdot 2 = - \frac{π}{15} \cdot 2

Simplify

\frac{3 π}{2} d \cdot 2 + 2 πn d \cdot 2 = - \frac{π}{15} \cdot 2

Simplify

\frac{3 π}{2} d \cdot 2 : 3 π d

\frac{3 π}{2} d \cdot 2

Multiply fractions:

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

= \frac{3 \cdot 2 π}{2} d

Cancel the common factor:

2

= d \cdot 3 π

Simplify

2 πn d \cdot 2 : 4 πn d

2 πn d \cdot 2

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn d

Simplify

- \frac{π}{15} \cdot 2 : - \frac{2 π}{15}

- \frac{π}{15} \cdot 2

Multiply fractions:

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

= - \frac{π 2}{15}

3 π d + 4 πn d = - \frac{2 π}{15}

3 π d + 4 πn d = - \frac{2 π}{15}

3 π d + 4 πn d = - \frac{2 π}{15}

Factor

3 π d + 4 πn d : π d (3 + 4 n)

3 π d + 4 πn d

Factor out common term

d π

= d π (3 + 4 n)

π d (3 + 4 n) = - \frac{2 π}{15}

Divide both sides by

π (3 + 4 n); n \neq = - \frac{3}{4}

π d (3 + 4 n) = - \frac{2 π}{15}

Divide both sides by

π (3 + 4 n); n \neq = - \frac{3}{4}

\frac{π d ( 3 + 4 n )}{π ( 3 + 4 n )} = \frac{- \frac{2 π}{15}}{π ( 3 + 4 n )}; n \neq = - \frac{3}{4}

Simplify

\frac{π d ( 3 + 4 n )}{π ( 3 + 4 n )} = \frac{- \frac{2 π}{15}}{π ( 3 + 4 n )}

Simplify

\frac{π d ( 3 + 4 n )}{π ( 3 + 4 n )} : d

\frac{π d ( 3 + 4 n )}{π ( 3 + 4 n )}

Cancel the common factor:

π

= \frac{d ( 4 n + 3 )}{3 + 4 n}

Cancel the common factor:

3 + 4 n

= d

Simplify

\frac{- \frac{2 π}{15}}{π ( 3 + 4 n )} : - \frac{2}{15 ( 4 n + 3 )}

\frac{- \frac{2 π}{15}}{π ( 3 + 4 n )}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{\frac{2 π}{15}}{π ( 3 + 4 n )}

Apply the fraction rule:

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

\frac{\frac{2 π}{15}}{π ( 3 + 4 n )} = \frac{2 π}{15 π ( 3 + 4 n )}

= - \frac{2 π}{15 π ( 3 + 4 n )}

Cancel the common factor:

π

= - \frac{2}{15 ( 4 n + 3 )}

d = - \frac{2}{15 ( 4 n + 3 )}; n \neq = - \frac{3}{4}

d = - \frac{2}{15 ( 4 n + 3 )}; n \neq = - \frac{3}{4}

d = - \frac{2}{15 ( 4 n + 3 )}; n \neq = - \frac{3}{4}

d = - \frac{2}{15 ( 4 n + 3 )}; n \neq = - \frac{3}{4}

Popular Examples

cos(θ)=(2sqrt(5))/(sqrt(58))

cos (θ) = \frac{2 5}{58}

tan(θ)=(m^2-m^1)/(1+m^2*m^1),m^2

tan (θ) = \frac{m ^{2} - m ^{1}}{1 + m ^{2} \cdot m ^{1}}, m^{2}

0=-1-tan^2(θ)+2tan(θ)

0 = - 1 - tan^{2} (θ) + 2 tan (θ)

solvefor u,x=tanh(u)solve for

u, x = tanh (u)

3+2cos(x)=0

3 + 2 cos (x) = 0

Frequently Asked Questions (FAQ)

What is the general solution for 15=60sin(pi/(15[0-d]))+75 ?
The general solution for 15=60sin(pi/(15[0-d]))+75 is d=-2/(15(4n+3));\quad n\ne-3/4