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

The general solution for 4sin(θ)=4(1-sin(θ)) is θ= pi/6+2pin,θ=(5pi)/6+2pin

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4sin(θ)=4(1-sin(θ))

4 sin (θ) = 4 (1 - 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 (1 - sin (θ))

Solve by substitution

4 sin (θ) = 4 (1 - sin (θ))

Let:

sin (θ) = u

4 u = 4 (1 - u)

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

4 u = 4 (1 - u)

Expand

4 (1 - u) : 4 - 4 u

4 (1 - u)

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = u

= 4 \cdot 1 - 4 u

Multiply the numbers:

4 \cdot 1 = 4

= 4 - 4 u

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

sin (2 x) = \frac{( 8 m - 2 )}{5}

csc (3 x) = sin (3 x)

arctan (x) = π + 1

sin^{2} (x) = \frac{( 10 m - 7 )}{9}

t, f = cos^{2} (a t)

What is the general solution for 4sin(θ)=4(1-sin(θ)) ?
The general solution for 4sin(θ)=4(1-sin(θ)) is θ= pi/6+2pin,θ=(5pi)/6+2pin