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

The general solution for 4sin(θ)=4-4sin(θ) is θ= pi/6+2pin,θ=(5pi)/6+2pin

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4sin(θ)=4-4sin(θ)

4 sin (θ) = 4 - 4 sin (θ)

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Degrees

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

4 sin (θ) = 4 - 4 sin (θ)

Solve by substitution

4 sin (θ) = 4 - 4 sin (θ)

Let:

sin (θ) = u

4 u = 4 - 4 u

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

4 u = 4 - 4 u

Move

4 u

to the left side

4 u = 4 - 4 u

Add

4 u

to both sides

4 u + 4 u = 4 - 4 u + 4 u

Simplify

8 u = 4

8 u = 4

Divide both sides by

8

8 u = 4

Divide both sides by

8

\frac{8 u}{8} = \frac{4}{8}

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{π}{6} + 2 πn, θ = \frac{5 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{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Combine all the solutions

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

tan (θ) = (- 3)

sin^{2} (x) + cos (x) + 1 = 0, 0 \leq x \leq 2 π

sin (2 x) = \frac{- 1}{3}

cos (2 θ) = 1 + 3 sin (θ)

cos (2 x) - cos (x) = - 1, 0 \leq x < 2 π

What is the general solution for 4sin(θ)=4-4sin(θ) ?
The general solution for 4sin(θ)=4-4sin(θ) is θ= pi/6+2pin,θ=(5pi)/6+2pin