What is the general solution for sin(θ)=-(sqrt(2))/2 ,0<= θ<2pi ?

The general solution for sin(θ)=-(sqrt(2))/2 ,0<= θ<2pi is θ=(5pi)/4 ,θ=(7pi)/4

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sin(θ)=-(sqrt(2))/2 ,0<= θ<2pi

sin (θ) = - \frac{2}{2}, 0 \leq θ < 2 π

θ = \frac{5 π}{4}, θ = \frac{7 π}{4}

Degrees

θ = 22 5^{\circ}, θ = 31 5^{\circ}

Solution steps

sin (θ) = - \frac{2}{2}, 0 \leq θ < 2 π

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

Solutions for the range

0 \leq θ < 2 π

θ = \frac{5 π}{4}, θ = \frac{7 π}{4}

2 sin (\frac{1}{2} x) + 2 = 0

\frac{cos ( x )}{sin ( x )} = - 0.05467

2 sin^{4} (x) + 2 sin^{2} (x) cos^{2} (x) = 1

(1 + 4 cos (θ))^{2} = (3 sin (θ))^{2}

\frac{1}{tan ( x )} = 4

What is the general solution for sin(θ)=-(sqrt(2))/2 ,0<= θ<2pi ?
The general solution for sin(θ)=-(sqrt(2))/2 ,0<= θ<2pi is θ=(5pi)/4 ,θ=(7pi)/4