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

The general solution for 0=2+2sin(θ) is θ=(3pi)/2+2pin

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0=2+2sin(θ)

0 = 2 + 2 sin (θ)

θ = \frac{3 π}{2} + 2 πn

Degrees

θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

0 = 2 + 2 sin (θ)

Switch sides

2 + 2 sin (θ) = 0

Move

2

to the right side

2 + 2 sin (θ) = 0

Subtract

2

from both sides

2 + 2 sin (θ) - 2 = 0 - 2

Simplify

2 sin (θ) = - 2

2 sin (θ) = - 2

Divide both sides by

2

2 sin (θ) = - 2

Divide both sides by

2

\frac{2 sin ( θ )}{2} = \frac{- 2}{2}

Simplify

sin (θ) = - 1

sin (θ) = - 1

General solutions for

sin (θ) = - 1

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{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

x, 2 cos (x) + 3 = 0

4 cos (x) - 5 = - 8

sin (x) + 4 csc (x) + 5 = 0, 0 \leq x \leq 2 π

r = a sin (3 x)

sin (x) = \frac{18}{25}

What is the general solution for 0=2+2sin(θ) ?
The general solution for 0=2+2sin(θ) is θ=(3pi)/2+2pin