Question

find all solutions $\theta$ to each of the following equations. separate multiple answers with commas. use none or not if there are no solutions. all angles are in radians.
$10s + 7sqrt{3}=2sqrt{3}$ has solutions $s = square$
$sin(\theta)=\frac{-sqrt{3}}{2}$ has solutions $\theta=square$
hint: dont forget your $2pi k$ for both - angles ($a,b + c$ means $a$ and then separately $b + c$).
webwork is a little too picky: $+2pi k$ is counted correct as expected, but $-2pi k$ is counted incorrect, even though technically it is correct, (just weird). webwork is ok with either $\frac{4pi}{3}$ or $-\frac{2pi}{3}$ - both are the same place on the circle.
$10sin(\theta)+7sqrt{3}=2sqrt{3}$ has solutions $\theta=square$

Explanation:

Step1: Solve for $S$ in $10S + 7\sqrt{3}=2\sqrt{3}$

Subtract $7\sqrt{3}$ from both sides:
$10S=2\sqrt{3}-7\sqrt{3}$
$10S=- 5\sqrt{3}$
Then divide both sides by 10:
$S =-\frac{\sqrt{3}}{2}$

Step2: Solve for $\theta$ in $\sin(\theta)=-\frac{\sqrt{3}}{2}$

We know that $\sin(\theta)=-\frac{\sqrt{3}}{2}$ in the third and fourth - quadrants.
The reference angle for $\sin$ value of $\frac{\sqrt{3}}{2}$ is $\frac{\pi}{3}$.
In the third - quadrant, $\theta=\pi+\frac{\pi}{3}=\frac{4\pi}{3}+2k\pi,k\in\mathbb{Z}$.
In the fourth - quadrant, $\theta = 2\pi-\frac{\pi}{3}=\frac{5\pi}{3}+2k\pi,k\in\mathbb{Z}$.

Step3: Solve for $\theta$ in $10\sin(\theta)+7\sqrt{3}=2\sqrt{3}$

First, subtract $7\sqrt{3}$ from both sides:
$10\sin(\theta)=2\sqrt{3}-7\sqrt{3}=-5\sqrt{3}$
Then divide both sides by 10:
$\sin(\theta)=-\frac{\sqrt{3}}{2}$
The solutions are $\theta=\frac{4\pi}{3}+2k\pi,\frac{5\pi}{3}+2k\pi,k\in\mathbb{Z}$

Answer:

$S =-\frac{\sqrt{3}}{2}$; $\theta=\frac{4\pi}{3}+2k\pi,\frac{5\pi}{3}+2k\pi$; $\theta=\frac{4\pi}{3}+2k\pi,\frac{5\pi}{3}+2k\pi$