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

The general solution for sin(θ)=(-4)/(4sqrt(2)),csc(θ)<0 is No Solution for θ\in\mathbb{R}

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sin(θ)=(-4)/(4sqrt(2)),csc(θ)<0

sin (θ) = \frac{- 4}{4 2}, csc (θ) < 0

No Solution for θ \in R

Solution steps

sin (θ) = \frac{- 4}{4 2}, csc (θ) < 0

Simplify

\frac{- 4}{4 2} : - \frac{2}{2}

\frac{- 4}{4 2}

Apply the fraction rule:

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

= - \frac{4}{4 2}

Divide the numbers:

\frac{4}{4} = 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

csc (θ) < 0

No Solution for θ \in R

sin^{2} (x) = 0.2

sin (x) = \frac{sin ( 4 5 ^{\circ} )}{1.3}

arctan (x) + arctan (1 - x) = arctan (\frac{9}{7})

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

3 cos^{2} (θ) - sin^{2} (θ) = 0

What is the general solution for sin(θ)=(-4)/(4sqrt(2)),csc(θ)<0 ?
The general solution for sin(θ)=(-4)/(4sqrt(2)),csc(θ)<0 is No Solution for θ\in\mathbb{R}