What is the general solution for 1+sin(θ)=2-sin(θ) ?

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

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1+sin(θ)=2-sin(θ)

1 + sin (θ) = 2 - sin (θ)

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

Solution steps

1 + sin (θ) = 2 - 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

- 4 cos^{2} (θ) + cos (θ) = 9 cos (θ) + 3

3 + 4 cot (x) = 5 csc (x)

(tan (x) - 1) (sec (x) + 1) = 0

tan (θ) = \frac{9}{10}

tan (x) = \frac{3}{205}

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