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

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

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

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

b = 2 πn, b = π + 2 πn

Degrees

b = 0^{\circ} + 36 0^{\circ} n, b = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

sin (b) = u

- 3 u - 2 = u - 2

- 3 u - 2 = u - 2 : u = 0

- 3 u - 2 = u - 2

Add

2

to both sides

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

Simplify

- 3 u = u

Move

u

to the left side

- 3 u = u

Subtract

u

from both sides

- 3 u - u = u - u

Simplify

- 4 u = 0

- 4 u = 0

Divide both sides by

- 4

- 4 u = 0

Divide both sides by

- 4

\frac{- 4 u}{- 4} = \frac{0}{- 4}

Simplify

u = 0

u = 0

Substitute back

u = sin (b)

sin (b) = 0

sin (b) = 0

sin (b) = 0 : b = 2 πn, b = π + 2 πn

sin (b) = 0

General solutions for

sin (b) = 0

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 = 0 + 2 πn, b = π + 2 πn

b = 0 + 2 πn, b = π + 2 πn

Solve

b = 0 + 2 πn : b = 2 πn

b = 0 + 2 πn

0 + 2 πn = 2 πn

b = 2 πn

b = 2 πn, b = π + 2 πn

Combine all the solutions

b = 2 πn, b = π + 2 πn

5 sin (2 x) = 2 cos (2 x)

tan (2 θ - 1 0^{\circ}) = cot (θ)

sinh (x) = 4

2 cos^{2} (θ) - 3 cos (θ) + 1 = 0, 0 \leq θ < 2 π

cos (x) sin (x) = 1

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