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

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

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(sin(θ)+1)(sin(θ)-1)=0

(sin (θ) + 1) (sin (θ) - 1) = 0

θ = \frac{3 π}{2} + 2 πn, θ = \frac{π}{2} + 2 πn

Degrees

θ = 27 0^{\circ} + 36 0^{\circ} n, θ = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

(sin (θ) + 1) (sin (θ) - 1) = 0

Solving each part separately

sin (θ) + 1 = 0 or sin (θ) - 1 = 0

sin (θ) + 1 = 0 : θ = \frac{3 π}{2} + 2 πn

sin (θ) + 1 = 0

Move

1

to the right side

sin (θ) + 1 = 0

Subtract

1

from both sides

sin (θ) + 1 - 1 = 0 - 1

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

sin (θ) - 1 = 0 : θ = \frac{π}{2} + 2 πn

sin (θ) - 1 = 0