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

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

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1-sin(θ)=sin(θ)

1 - sin (θ) = 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

1 - sin (θ) = sin (θ)

Solve by substitution

1 - sin (θ) = sin (θ)

Let:

sin (θ) = u

1 - u = u

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

1 - u = u

Move

1

to the right side

1 - u = u

Subtract

1

from both sides

1 - u - 1 = u - 1

Simplify

- u = u - 1

- u = u - 1

Move

u

to the left side

- u = u - 1

Subtract

u

from both sides

- u - u = u - 1 - 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

cot (\frac{θ}{3} + \frac{3 π}{4}) = 3

cos (x) = \frac{tan ( x ) + cot ( x )}{csc ( x )}

- tan^{2} (x) - 1 = 0

2 cot^{2} (y) + 3 csc (y) = 0

cot (x) = - 0.456

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