What is the general solution for sin(θ)=-(sqrt(2))/2 ?

The general solution for sin(θ)=-(sqrt(2))/2 is θ=(5pi)/4+2pin,θ=(7pi)/4+2pin

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sin(θ)=-(sqrt(2))/2

sin (θ) = - \frac{2}{2}

θ = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

Degrees

θ = 22 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n

Solution steps

sin (θ) = - \frac{2}{2}

General solutions for

sin (θ) = - \frac{2}{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{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

θ = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

cos^{2} (0)

sech (0)

cos (2 θ) = 0

sin (4 π)

sin (\frac{7 π}{3})

What is the general solution for sin(θ)=-(sqrt(2))/2 ?
The general solution for sin(θ)=-(sqrt(2))/2 is θ=(5pi)/4+2pin,θ=(7pi)/4+2pin