Solve for θ in Degrees csc(theta)=-(2 square root of 3)/3

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Trigonometry

$\csc (θ) = - \frac{2 \sqrt{3}}{3}$

Take the inverse cosecant of both sides of the equation to extract $θ$ from inside the cosecant.

$θ = arccsc (- \frac{2 \sqrt{3}}{3})$

Simplify the right side.

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The exact value of $arccsc (- \frac{2 \sqrt{3}}{3})$ is $- 60$ .

$θ = - 60$

The cosecant function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $360$ , to find a reference angle. Next, add this reference angle to $180$ to find the solution in the third quadrant.

$θ = 360 + 60 + 180$

Simplify the expression to find the second solution.

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Subtract $360 °$ from $360 + 60 + 180 °$ .

$θ = 360 + 60 + 180 ° - 360 °$

The resulting angle of $240 °$ is positive, less than $360 °$ , and coterminal with $360 + 60 + 180$ .

$θ = 240 °$

Find the period of $\csc (θ)$ .

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The period of the function can be calculated using $\frac{360}{| b |}$ .

$\frac{360}{| b |}$

Replace $b$ with $1$ in the formula for period.

$\frac{360}{| 1 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{360}{1}$

Divide $360$ by $1$ .

$360$

Add $360$ to every negative angle to get positive angles.

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Add $360$ to $- 60$ to find the positive angle.

$- 60 + 360$

Subtract $60$ from $360$ .

$300$

List the new angles.

$θ = 300$

The period of the $\csc (θ)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$θ = 240 + 360 n, 300 + 360 n$ , for any integer $n$

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