What is the general solution for 1+2sin(θ)=0 ?

The general solution for 1+2sin(θ)=0 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin

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1+2sin(θ)=0

1 + 2 sin (θ) = 0

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Solution steps

1 + 2 sin (θ) = 0

Move

1

to the right side

2 sin (θ) = - 1

Divide both sides by

2

sin (θ) = - \frac{1}{2}

General solutions for

sin (θ) = - \frac{1}{2}

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

x, z = arcsin (x y)

2 cos (x) - 2 = 0

4 sin^{2} (θ) - 13 sin (θ) + 9 = 0

23 cos^{2} (x) = sin (x)

4 cos^{4} (θ) - 5 cos^{2} (θ) + 1 = 0

What is the general solution for 1+2sin(θ)=0 ?
The general solution for 1+2sin(θ)=0 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin