What is the general solution for 7=10sin(40(0-h))+17 ?

The general solution for 7=10sin(40(0-h))+17 is h=-(3pi)/(80)-(pin)/(20)

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Popular Trigonometry >

7=10sin(40(0-h))+17

Solution

7 = 10 sin (40 (0 - h)) + 17

Solution

h = - \frac{3 π}{80} - \frac{πn}{20}

Degrees

h = - 6.7 5^{\circ} - 9^{\circ} n

Solution steps

7 = 10 sin (40 (0 - h)) + 17

Switch sides

10 sin (40 (0 - h)) + 17 = 7

Move

17

to the right side

10 sin (40 (0 - h)) + 17 = 7

Subtract

17

from both sides

10 sin (40 (0 - h)) + 17 - 17 = 7 - 17

Simplify

10 sin (40 (0 - h)) = - 10

10 sin (40 (0 - h)) = - 10

Divide both sides by

10

10 sin (40 (0 - h)) = - 10

Divide both sides by

10

\frac{10 sin ( 40 ( 0 - h ) )}{10} = \frac{- 10}{10}

Simplify

sin (40 (0 - h)) = - 1

sin (40 (0 - h)) = - 1

General solutions for

sin (40 (0 - h)) = - 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}

40 (0 - h) = \frac{3 π}{2} + 2 πn

40 (0 - h) = \frac{3 π}{2} + 2 πn

Solve

40 (0 - h) = \frac{3 π}{2} + 2 πn : h = - \frac{3 π}{80} - \frac{πn}{20}

40 (0 - h) = \frac{3 π}{2} + 2 πn

Divide both sides by

40

40 (0 - h) = \frac{3 π}{2} + 2 πn

Divide both sides by

40

\frac{40 ( 0 - h )}{40} = \frac{\frac{3 π}{2}}{40} + \frac{2 πn}{40}

Simplify

\frac{40 ( 0 - h )}{40} = \frac{\frac{3 π}{2}}{40} + \frac{2 πn}{40}

Simplify

\frac{40 ( 0 - h )}{40} : - h

\frac{40 ( 0 - h )}{40}

40 (0 - h) = - 40 h

40 (0 - h)

0 - h = - h

= 40 (- h)

Remove parentheses:

(- a) = - a

= - 40 h

= \frac{- 40 h}{40}

Apply the fraction rule:

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

= - \frac{40 h}{40}

Divide the numbers:

\frac{40}{40} = 1

= - h

Simplify

\frac{\frac{3 π}{2}}{40} + \frac{2 πn}{40} : \frac{3 π}{80} + \frac{πn}{20}

\frac{\frac{3 π}{2}}{40} + \frac{2 πn}{40}

\frac{\frac{3 π}{2}}{40} = \frac{3 π}{80}

\frac{\frac{3 π}{2}}{40}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 40}

Multiply the numbers:

2 \cdot 40 = 80

= \frac{3 π}{80}

\frac{2 πn}{40} = \frac{πn}{20}

\frac{2 πn}{40}

Cancel the common factor:

2

= \frac{πn}{20}

= \frac{3 π}{80} + \frac{πn}{20}

- h = \frac{3 π}{80} + \frac{πn}{20}

- h = \frac{3 π}{80} + \frac{πn}{20}

- h = \frac{3 π}{80} + \frac{πn}{20}

Divide both sides by

- 1

- h = \frac{3 π}{80} + \frac{πn}{20}

Divide both sides by

- 1

\frac{- h}{- 1} = \frac{\frac{3 π}{80}}{- 1} + \frac{\frac{πn}{20}}{- 1}

Simplify

\frac{- h}{- 1} = \frac{\frac{3 π}{80}}{- 1} + \frac{\frac{πn}{20}}{- 1}

Simplify

\frac{- h}{- 1} : h

\frac{- h}{- 1}

Apply the fraction rule:

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

= \frac{h}{1}

Apply rule

\frac{a}{1} = a

= h

Simplify

\frac{\frac{3 π}{80}}{- 1} + \frac{\frac{πn}{20}}{- 1} : - \frac{3 π}{80} - \frac{πn}{20}

\frac{\frac{3 π}{80}}{- 1} + \frac{\frac{πn}{20}}{- 1}

\frac{\frac{3 π}{80}}{- 1} = - \frac{3 π}{80}

\frac{\frac{3 π}{80}}{- 1}

Apply the fraction rule:

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

= - \frac{\frac{3 π}{80}}{1}

Apply the fraction rule:

\frac{a}{1} = a

\frac{\frac{3 π}{80}}{1} = \frac{3 π}{80}

= - \frac{3 π}{80}

= - \frac{3 π}{80} + \frac{\frac{πn}{20}}{- 1}

\frac{\frac{πn}{20}}{- 1} = - \frac{πn}{20}

\frac{\frac{πn}{20}}{- 1}

Apply the fraction rule:

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

= - \frac{\frac{πn}{20}}{1}

Apply the fraction rule:

\frac{a}{1} = a

\frac{\frac{πn}{20}}{1} = \frac{πn}{20}

= - \frac{πn}{20}

= - \frac{3 π}{80} - \frac{πn}{20}

h = - \frac{3 π}{80} - \frac{πn}{20}

h = - \frac{3 π}{80} - \frac{πn}{20}

h = - \frac{3 π}{80} - \frac{πn}{20}

h = - \frac{3 π}{80} - \frac{πn}{20}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 7=10sin(40(0-h))+17 ?
The general solution for 7=10sin(40(0-h))+17 is h=-(3pi)/(80)-(pin)/(20)