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

Degrees

θ = 21 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

1 + 2 sin (θ) = 0

Move

1

to the right side

1 + 2 sin (θ) = 0

Subtract

1

from both sides

1 + 2 sin (θ) - 1 = 0 - 1

Simplify

2 sin (θ) = - 1

2 sin (θ) = - 1

Divide both sides by

2

2 sin (θ) = - 1

Divide both sides by

2

\frac{2 sin ( θ )}{2} = \frac{- 1}{2}

Simplify

sin (θ) = - \frac{1}{2}

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

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