What is the general solution for 9sin(b)+3=5sin(b)+1 ?

The general solution for 9sin(b)+3=5sin(b)+1 is b=(7pi)/6+2pin,b=(11pi)/6+2pin

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9sin(b)+3=5sin(b)+1

Solution

9 sin (b) + 3 = 5 sin (b) + 1

Solution

b = \frac{7 π}{6} + 2 πn, b = \frac{11 π}{6} + 2 πn

Degrees

b = 21 0^{\circ} + 36 0^{\circ} n, b = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

9 sin (b) + 3 = 5 sin (b) + 1

Solve by substitution

9 sin (b) + 3 = 5 sin (b) + 1

Let:

sin (b) = u

9 u + 3 = 5 u + 1

9 u + 3 = 5 u + 1 : u = - \frac{1}{2}

9 u + 3 = 5 u + 1

Move

3

to the right side

9 u + 3 = 5 u + 1

Subtract

3

from both sides

9 u + 3 - 3 = 5 u + 1 - 3

Simplify

9 u = 5 u - 2

9 u = 5 u - 2

Move

5 u

to the left side

9 u = 5 u - 2

Subtract

5 u

from both sides

9 u - 5 u = 5 u - 2 - 5 u

Simplify

4 u = - 2

4 u = - 2

Divide both sides by

4

4 u = - 2

Divide both sides by

4

\frac{4 u}{4} = \frac{- 2}{4}

Simplify

\frac{4 u}{4} = \frac{- 2}{4}

Simplify

\frac{4 u}{4} : u

\frac{4 u}{4}

Divide the numbers:

\frac{4}{4} = 1

= u

Simplify

\frac{- 2}{4} : - \frac{1}{2}

\frac{- 2}{4}

Apply the fraction rule:

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

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

u = - \frac{1}{2}

u = - \frac{1}{2}

u = - \frac{1}{2}

Substitute back

u = sin (b)

sin (b) = - \frac{1}{2}

sin (b) = - \frac{1}{2}

sin (b) = - \frac{1}{2} : b = \frac{7 π}{6} + 2 πn, b = \frac{11 π}{6} + 2 πn

sin (b) = - \frac{1}{2}

General solutions for

sin (b) = - \frac{1}{2}

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}

b = \frac{7 π}{6} + 2 πn, b = \frac{11 π}{6} + 2 πn

b = \frac{7 π}{6} + 2 πn, b = \frac{11 π}{6} + 2 πn

Combine all the solutions

b = \frac{7 π}{6} + 2 πn, b = \frac{11 π}{6} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 9sin(b)+3=5sin(b)+1 ?
The general solution for 9sin(b)+3=5sin(b)+1 is b=(7pi)/6+2pin,b=(11pi)/6+2pin