What is the general solution for 3sin(b)+3=5sin(b)+2 ?

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

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3sin(b)+3=5sin(b)+2

3 sin (b) + 3 = 5 sin (b) + 2

b = \frac{π}{6} + 2 πn, b = \frac{5 π}{6} + 2 πn

Degrees

b = 3 0^{\circ} + 36 0^{\circ} n, b = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

3 sin (b) + 3 = 5 sin (b) + 2

Solve by substitution

3 sin (b) + 3 = 5 sin (b) + 2

Let:

sin (b) = u

3 u + 3 = 5 u + 2

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

3 u + 3 = 5 u + 2

Move

3

to the right side

3 u + 3 = 5 u + 2

Subtract

3

from both sides

3 u + 3 - 3 = 5 u + 2 - 3

Simplify

3 u = 5 u - 1

3 u = 5 u - 1

Move

5 u

to the left side

3 u = 5 u - 1

Subtract

5 u

from both sides

3 u - 5 u = 5 u - 1 - 5 u

Simplify

- 2 u = - 1

- 2 u = - 1

Divide both sides by

- 2

- 2 u = - 1

Divide both sides by

- 2

\frac{- 2 u}{- 2} = \frac{- 1}{- 2}

Simplify

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{π}{6} + 2 πn, b = \frac{5 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{π}{6} + 2 πn, b = \frac{5 π}{6} + 2 πn

b = \frac{π}{6} + 2 πn, b = \frac{5 π}{6} + 2 πn

Combine all the solutions

b = \frac{π}{6} + 2 πn, b = \frac{5 π}{6} + 2 πn

8 tan^{2} (θ) + 10 tan (θ) + 10 = 7

3 sec (x) - 23 = 0

2 cos (x) - 3 = - 5

sin (3 θ) - sin (θ) = cos (2 θ)

2 sin (x) + (2 - 3) = 3 csc (x)

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