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

Solution steps

sin (θ) = 1 - sin (θ)

Solve by substitution

sin (θ) = \frac{1}{2}

sin (θ) = \frac{1}{2} : 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