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

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

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sin(θ)=(-2)/(2sqrt(2)),0<= θ<2pi

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

θ = \frac{5 π}{4}, θ = \frac{7 π}{4}

Solution steps

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

Simplify

\frac{- 2}{2 2} : - \frac{2}{2}

\frac{- 2}{2 2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2}{2 2}

Divide the numbers:

\frac{2}{2} = 1

= - \frac{1}{2}

Rationalize

- \frac{1}{2} : - \frac{2}{2}

- \frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= - \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

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

Solutions for the range

0 \leq θ < 2 π

θ = \frac{5 π}{4}, θ = \frac{7 π}{4}

(sin (θ) + 1) (sin (θ) - 1) = 0

\frac{30}{sin ( x )} = \frac{50}{sin ( 9 0 ^{\circ} )}

cos (θ) = \frac{2 3}{5}

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

5 sin (θ) = 5

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