Find all solutions to $2 \sin (θ) = - \sqrt{3}$ on the interval $0 \leq θ < 2 π$
$θ =$

Give your answers as exact values, as a list separated by commas.
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The solutions to $2 \sin (θ) = - \sqrt{3}$ on the interval $0 \leq θ < 2 π$ are $θ = \frac{4 π}{3}, \frac{5 π}{3}$ .

Steps

Step 1 ：Given the equation is $2 \sin (θ) = - \sqrt{3}$ . We need to find all solutions for $θ$ in the interval $0 \leq θ < 2 π$ .

Step 2 ：Simplify the equation to $\sin (θ) = - \frac{\sqrt{3}}{2}$ .

Step 3 ： $\sin (θ)$ is negative in the third and fourth quadrants.

Step 4 ：The reference angle for $\frac{\sqrt{3}}{2}$ is $\frac{π}{3}$ or $60^{\circ}$ .

Step 5 ：The solutions in the third and fourth quadrants would be $π + \frac{π}{3}$ and $2 π - \frac{π}{3}$ respectively.

Step 6 ：The exact values of $θ$ in the third and fourth quadrants would be $\frac{4 π}{3}$ and $\frac{5 π}{3}$ respectively.

Step 7 ：The solutions to $2 \sin (θ) = - \sqrt{3}$ on the interval $0 \leq θ < 2 π$ are $θ = \frac{4 π}{3}, \frac{5 π}{3}$ .