How do you find the values of $θ$ $theta$ given $csc θ = \frac{2 \sqrt{3}}{3}$ $csctheta=(2sqrt3)/3$ ?

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Answer 1 · 2017-03-21T15:00:46

$θ = n π \pm {(- 1)}^{n} \frac{π}{3}$ $theta=npi+-(-1)^npi/3$ , where $n$ $n$ is an integer.

As $csc θ = \frac{2 \sqrt{3}}{3} = \frac{2}{\sqrt{3}}$ $csctheta=(2sqrt3)/3=2/sqrt3$

Therefore, $sin θ = \frac{\sqrt{3}}{2} = sin (\frac{π}{3})$ $sintheta=sqrt3/2=sin(pi/3)$

Hence, within the interval $0 \leq θ \leq 2 π$ $0 <= theta <= 2pi$ , $θ = \frac{π}{3}$ $theta=pi/3$ or $\frac{2 π}{3}$ $(2pi)/3$

but as trigonometric ratios repeat after $2 π$ $2pi$

they are more solutions which can be put as

$θ = n π \pm {(- 1)}^{n} \frac{π}{3}$ $theta=npi+-(-1)^npi/3$ , where $n$ $n$ is an integer.

How do you find the values of θtheta given cscθ=2√33csctheta=(2sqrt3)/3?