What is the general solution for solvefor x,sin(x)+1=0 ?

The general solution for solvefor x,sin(x)+1=0 is x=(3pi)/2+2pin

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solvefor x,sin(x)+1=0

solve for

x, sin (x) + 1 = 0

x = \frac{3 π}{2} + 2 πn

Degrees

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

Solution steps

sin (x) + 1 = 0

Move

1

to the right side

sin (x) + 1 = 0

Subtract

1

from both sides

sin (x) + 1 - 1 = 0 - 1

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

\frac{8}{sin ( 3 8 ^{\circ} )} = \frac{6}{sin ( a )}

28 cos (θ) = 4, 0^{\circ} \leq θ < 36 0^{\circ}

6 sin (θ) + 2 = sin (θ)

cos (θ) = - \frac{19}{17}

5 = 7 - cos (θ) + 5 cos (2 θ)

What is the general solution for solvefor x,sin(x)+1=0 ?
The general solution for solvefor x,sin(x)+1=0 is x=(3pi)/2+2pin