Solve for θ square root of 3csc(theta)+2=0

$\sqrt{3} \csc (θ) + 2 = 0$

Subtract $2$ from both sides of the equation.

$\sqrt{3} \csc (θ) = - 2$

Divide each term in $\sqrt{3} \csc (θ) = - 2$ by $\sqrt{3}$ and simplify.

Divide each term in $\sqrt{3} \csc (θ) = - 2$ by $\sqrt{3}$ .

$\frac{\sqrt{3} \csc (θ)}{\sqrt{3}} = \frac{- 2}{\sqrt{3}}$

Simplify the left side.

Cancel the common factor of $\sqrt{3}$ .

Cancel the common factor.

$\frac{\sqrt{3} \csc (θ)}{\sqrt{3}} = \frac{- 2}{\sqrt{3}}$

Divide $\csc (θ)$ by $1$ .

$\csc (θ) = \frac{- 2}{\sqrt{3}}$

Simplify the right side.

Move the negative in front of the fraction.

$\csc (θ) = - \frac{2}{\sqrt{3}}$

Multiply $\frac{2}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ .

$\csc (θ) = - (\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}})$

Combine and simplify the denominator.

Multiply $\frac{2}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ .

$\csc (θ) = - \frac{2 \sqrt{3}}{\sqrt{3} \sqrt{3}}$

Raise $\sqrt{3}$ to the power of $1$ .

$\csc (θ) = - \frac{2 \sqrt{3}}{{\sqrt{3}}^{1} \sqrt{3}}$

Raise $\sqrt{3}$ to the power of $1$ .

$\csc (θ) = - \frac{2 \sqrt{3}}{{\sqrt{3}}^{1} {\sqrt{3}}^{1}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\csc (θ) = - \frac{2 \sqrt{3}}{{\sqrt{3}}^{1 + 1}}$

Add $1$ and $1$ .

$\csc (θ) = - \frac{2 \sqrt{3}}{{\sqrt{3}}^{2}}$

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$\csc (θ) = - \frac{2 \sqrt{3}}{{(3^{\frac{1}{2}})}^{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\csc (θ) = - \frac{2 \sqrt{3}}{3^{\frac{1}{2} \cdot 2}}$

Combine $\frac{1}{2}$ and $2$ .

$\csc (θ) = - \frac{2 \sqrt{3}}{3^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\csc (θ) = - \frac{2 \sqrt{3}}{3^{\frac{2}{2}}}$

Rewrite the expression.

$\csc (θ) = - \frac{2 \sqrt{3}}{3^{1}}$

Evaluate the exponent.

$\csc (θ) = - \frac{2 \sqrt{3}}{3}$

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.

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

$θ = - \frac{π}{3}$

The cosecant function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2 π$ , to find a reference angle. Next, add this reference angle to $π$ to find the solution in the third quadrant.

$θ = 2 π + \frac{π}{3} + π$

Simplify the expression to find the second solution.

Subtract $2 π$ from $2 π + \frac{π}{3} + π$ .

$θ = 2 π + \frac{π}{3} + π - 2 π$

The resulting angle of $\frac{4 π}{3}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{3} + π$ .

$θ = \frac{4 π}{3}$

Find the period of $\csc (θ)$ .

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

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

$\frac{2 π}{| 1 |}$

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

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

Add $2 π$ to every negative angle to get positive angles.

Add $2 π$ to $- \frac{π}{3}$ to find the positive angle.

$- \frac{π}{3} + 2 π$

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$2 π \cdot \frac{3}{3} - \frac{π}{3}$

Combine fractions.

Combine $2 π$ and $\frac{3}{3}$ .

$\frac{2 π \cdot 3}{3} - \frac{π}{3}$

Combine the numerators over the common denominator.

$\frac{2 π \cdot 3 - π}{3}$

Simplify the numerator.

Multiply $3$ by $2$ .

$\frac{6 π - π}{3}$

Subtract $π$ from $6 π$ .

$\frac{5 π}{3}$

List the new angles.

$θ = \frac{5 π}{3}$

The period of the $\csc (θ)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$θ = \frac{4 π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

Solve for θ square root of 3csc(theta)+2=0

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