What is the general solution for 11sin(θ)+2=4sin(θ)+2 ?

The general solution for 11sin(θ)+2=4sin(θ)+2 is θ=2pin,θ=pi+2pin

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11sin(θ)+2=4sin(θ)+2

11 sin (θ) + 2 = 4 sin (θ) + 2

θ = 2 πn, θ = π + 2 πn

Degrees

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

Solution steps

11 sin (θ) + 2 = 4 sin (θ) + 2

Solve by substitution

11 sin (θ) + 2 = 4 sin (θ) + 2

Let:

sin (θ) = u

11 u + 2 = 4 u + 2

11 u + 2 = 4 u + 2 : u = 0

11 u + 2 = 4 u + 2

Subtract

2

from both sides

11 u + 2 - 2 = 4 u + 2 - 2

Simplify

11 u = 4 u

Move

4 u

to the left side

11 u = 4 u

Subtract

4 u

from both sides

11 u - 4 u = 4 u - 4 u

Simplify

7 u = 0

7 u = 0

Divide both sides by

7

7 u = 0

Divide both sides by

7

\frac{7 u}{7} = \frac{0}{7}

Simplify

u = 0

u = 0

Substitute back

u = sin (θ)

sin (θ) = 0

sin (θ) = 0

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = 0

General solutions for

sin (θ) = 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}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn

sec^{2} (x) - tan (x) = 1

cos (θ) = \frac{11}{130}

2 sin^{2} (u) = 1 - sin (u)

2 cos^{2} (x) = 1 - sin (x)

sin (t) = - \frac{3}{2}

What is the general solution for 11sin(θ)+2=4sin(θ)+2 ?
The general solution for 11sin(θ)+2=4sin(θ)+2 is θ=2pin,θ=pi+2pin