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

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

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2+2sin(x)=0

2 + 2 sin (x) = 0

x = \frac{3 π}{2} + 2 πn

Degrees

x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

2 + 2 sin (x) = 0

Move

2

to the right side

2 + 2 sin (x) = 0

Subtract

2

from both sides

2 + 2 sin (x) - 2 = 0 - 2

Simplify

2 sin (x) = - 2

2 sin (x) = - 2

Divide both sides by

2

2 sin (x) = - 2

Divide both sides by

2

\frac{2 sin ( x )}{2} = \frac{- 2}{2}

Simplify

sin (x) = - 1

sin (x) = - 1

General solutions for

sin (x) = - 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}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

2 sin (2 x) = 1, 0 < x < 2 π

sin^{2} (\frac{x}{2}) = cos (x) - 1

sin (x) = cos (2 x), 0 \leq x \leq 2 π

sin (2 x) = sin (x) cos (x)

sin (θ) = - \frac{5}{5}, \frac{3 π}{2} < θ < 2 π

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