What is the general solution for 2sin(θ)=-sqrt(3)+4sin(θ) ?

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

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2sin(θ)=-sqrt(3)+4sin(θ)

2 sin (θ) = - 3 + 4 sin (θ)

θ = \frac{π}{3} + 2 πn, θ = \frac{2 π}{3} + 2 πn

Degrees

θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 12 0^{\circ} + 36 0^{\circ} n

Solution steps

2 sin (θ) = - 3 + 4 sin (θ)

Solve by substitution

2 sin (θ) = - 3 + 4 sin (θ)

Let:

sin (θ) = u

2 u = - 3 + 4 u

2 u = - 3 + 4 u : u = \frac{3}{2}

2 u = - 3 + 4 u

Move

4 u

to the left side

2 u = - 3 + 4 u

Subtract

4 u

from both sides

2 u - 4 u = - 3 + 4 u - 4 u

Simplify

- 2 u = - 3

- 2 u = - 3

Divide both sides by

- 2

- 2 u = - 3

Divide both sides by

- 2

\frac{- 2 u}{- 2} = \frac{- 3}{- 2}

Simplify

u = \frac{3}{2}

u = \frac{3}{2}

Substitute back

u = sin (θ)

sin (θ) = \frac{3}{2}

sin (θ) = \frac{3}{2}

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

sin (θ) = \frac{3}{2}

General solutions for

sin (θ) = \frac{3}{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{π}{3} + 2 πn, θ = \frac{2 π}{3} + 2 πn

θ = \frac{π}{3} + 2 πn, θ = \frac{2 π}{3} + 2 πn

Combine all the solutions

θ = \frac{π}{3} + 2 πn, θ = \frac{2 π}{3} + 2 πn

7^{2} = 6^{2} + 1 1^{2} - 2 \cdot 6 \cdot 11 \cdot cos (D)

- 4 cos^{2} (x) + 3 cos (x) + 1 = 0

cos^{2} (θ) + \frac{5 ^{2} cos ( θ )}{9.81 ( 8 )} - 1 = 0

105 = 100 + 30 sin (\frac{12 π}{5} t)

\frac{3}{sin ( x )} = \frac{1}{cos ( x )}

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