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

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

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sin(θ)=1-sin(θ)

sin (θ) = 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

sin (θ) = 1 - sin (θ)

Solve by substitution

sin (θ) = 1 - sin (θ)

Let:

sin (θ) = u

u = 1 - u

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

u = 1 - u

Move

u

to the left side

u = 1 - u

Add

u

to both sides

u + u = 1 - u + 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{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

2 sin (x) - 3 = 0, 0 \leq x \leq 2 π

cos (θ) = cot (θ)

8 cos (x) + 4 sin (2 x) = 0

sin (x) = - \frac{4}{5}, cos (x) < 0, cos (\frac{x}{2})

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

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