What is the general solution for 4sin(θ)+1=2sin(θ) ?

The general solution for 4sin(θ)+1=2sin(θ) is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin

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4sin(θ)+1=2sin(θ)

4 sin (θ) + 1 = 2 sin (θ)

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Degrees

θ = 21 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

4 sin (θ) + 1 = 2 sin (θ)

Solve by substitution

4 sin (θ) + 1 = 2 sin (θ)

Let:

sin (θ) = u

4 u + 1 = 2 u

4 u + 1 = 2 u : u = - \frac{1}{2}

4 u + 1 = 2 u

Move

1

to the right side

4 u + 1 = 2 u

Subtract

1

from both sides

4 u + 1 - 1 = 2 u - 1

Simplify

4 u = 2 u - 1

4 u = 2 u - 1

Move

2 u

to the left side

4 u = 2 u - 1

Subtract

2 u

from both sides

4 u - 2 u = 2 u - 1 - 2 u

Simplify

2 u = - 1

2 u = - 1

Divide both sides by

2

2 u = - 1

Divide both sides by

2

\frac{2 u}{2} = \frac{- 1}{2}

Simplify

u = - \frac{1}{2}

u = - \frac{1}{2}

Substitute back

u = sin (θ)

sin (θ) = - \frac{1}{2}

sin (θ) = - \frac{1}{2}

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

sin (θ) = - \frac{1}{2}

General solutions for

sin (θ) = - \frac{1}{2}

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{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Combine all the solutions

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

cot (θ) = \frac{1}{2}

sin (θ) = 0.7

sin (θ) = 0.2

sin^{2} (x) - 2 sin (x) + 1 = 0

2 sin (x) = - 1

What is the general solution for 4sin(θ)+1=2sin(θ) ?
The general solution for 4sin(θ)+1=2sin(θ) is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin