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

The general solution for 3sin(θ)+5=-2sin(θ) is θ=(3pi)/2+2pin

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3sin(θ)+5=-2sin(θ)

3 sin (θ) + 5 = - 2 sin (θ)

θ = \frac{3 π}{2} + 2 πn

Degrees

θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

3 sin (θ) + 5 = - 2 sin (θ)

Solve by substitution

3 sin (θ) + 5 = - 2 sin (θ)

Let:

sin (θ) = u

3 u + 5 = - 2 u

3 u + 5 = - 2 u : u = - 1

3 u + 5 = - 2 u

Move

5

to the right side

3 u + 5 = - 2 u

Subtract

5

from both sides

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

Simplify

3 u = - 2 u - 5

3 u = - 2 u - 5

Move

2 u

to the left side

3 u = - 2 u - 5

Add

2 u

to both sides

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

Simplify

5 u = - 5

5 u = - 5

Divide both sides by

5

5 u = - 5

Divide both sides by

5

\frac{5 u}{5} = \frac{- 5}{5}

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

2 - 4 sin (θ) = 0

2 cos (t) = 1, - π \leq t \leq π

tan (\frac{x}{3}) = 1, 0 \leq x \leq 2 π

4 sin (x) = sin (x) tan (x)

cos (x) = 0.707

What is the general solution for 3sin(θ)+5=-2sin(θ) ?
The general solution for 3sin(θ)+5=-2sin(θ) is θ=(3pi)/2+2pin