What is the general solution for -2sin(b)-6=sin(b)-3 ?

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

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-2sin(b)-6=sin(b)-3

- 2 sin (b) - 6 = sin (b) - 3

b = \frac{3 π}{2} + 2 πn

Degrees

b = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

- 2 sin (b) - 6 = sin (b) - 3

Solve by substitution

- 2 sin (b) - 6 = sin (b) - 3

Let:

sin (b) = u

- 2 u - 6 = u - 3

- 2 u - 6 = u - 3 : u = - 1

- 2 u - 6 = u - 3

Move

6

to the right side

- 2 u - 6 = u - 3

Add

6

to both sides

- 2 u - 6 + 6 = u - 3 + 6

Simplify

- 2 u = u + 3

- 2 u = u + 3

Move

u

to the left side

- 2 u = u + 3

Subtract

u

from both sides

- 2 u - u = u + 3 - u

Simplify

- 3 u = 3

- 3 u = 3

Divide both sides by

- 3

- 3 u = 3

Divide both sides by

- 3

\frac{- 3 u}{- 3} = \frac{3}{- 3}

Simplify

u = - 1

u = - 1

Substitute back

u = sin (b)

sin (b) = - 1

sin (b) = - 1

sin (b) = - 1 : b = \frac{3 π}{2} + 2 πn

sin (b) = - 1

General solutions for

sin (b) = - 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}

b = \frac{3 π}{2} + 2 πn

b = \frac{3 π}{2} + 2 πn

Combine all the solutions

b = \frac{3 π}{2} + 2 πn

cos (t) = - sin (t)

cos (4 x) \cdot 4 = 0

cos^{2} (x) = \frac{3}{2}

sin (\frac{x}{2}) = \frac{3}{5}

16 cos^{2} (x) - 8 = 0

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