What is the general solution for -6sin(θ)-3=sin(θ)+4 ?

The general solution for -6sin(θ)-3=sin(θ)+4 is θ=(3pi)/2+2pin

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-6sin(θ)-3=sin(θ)+4

- 6 sin (θ) - 3 = sin (θ) + 4

θ = \frac{3 π}{2} + 2 πn

Degrees

θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

- 6 sin (θ) - 3 = sin (θ) + 4

Solve by substitution

- 6 sin (θ) - 3 = sin (θ) + 4

Let:

sin (θ) = u

- 6 u - 3 = u + 4

- 6 u - 3 = u + 4 : u = - 1

- 6 u - 3 = u + 4

Move

3

to the right side

- 6 u - 3 = u + 4

Add

3

to both sides

- 6 u - 3 + 3 = u + 4 + 3

Simplify

- 6 u = u + 7

- 6 u = u + 7

Move

u

to the left side

- 6 u = u + 7

Subtract

u

from both sides

- 6 u - u = u + 7 - u

Simplify

- 7 u = 7

- 7 u = 7

Divide both sides by

- 7

- 7 u = 7

Divide both sides by

- 7

\frac{- 7 u}{- 7} = \frac{7}{- 7}

Simplify

u = - 1

u = - 1

Substitute back

u = sin (θ)

sin (θ) = - 1

sin (θ) = - 1

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

sin (θ) = - 1

General solutions for

sin (θ) = - 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}

θ = \frac{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

Combine all the solutions

θ = \frac{3 π}{2} + 2 πn

- 9 cos^{2} (θ) + 23 sin (θ) + 17 = - 2

24 arcsin (x) = 4 π

cos (x) - 1 = sin (x)

2 cos^{2} (x) + 3 sin (x) - 3 = 0, 0^{\circ} \leq x \leq 36 0^{\circ}

5 cot (θ) + 7 = 0

What is the general solution for -6sin(θ)-3=sin(θ)+4 ?
The general solution for -6sin(θ)-3=sin(θ)+4 is θ=(3pi)/2+2pin