Question

hw07b fall25: problem 7 (1 point) find all solutions $\theta$ to each of the following equations. separate multiple answers with commas. use solutions. all angles are in radians. 10s + 7 = 5\sqrt{3}+7 has solutions s = sin ($\theta$) = \frac{\sqrt{3}}{2} has solutions $\theta$ = hint: dont forget your 2$\pi$k for both angles (a, b + c means a and then separately b + c) 10 sin ($\theta$)+7 = 5\sqrt{3}+7 has solutions $\theta$ = note: you can earn partial credit on this problem. preview my answers submit answers you have attempted this problem 0 times. you have unlimited attempts remaining.

Explanation:

Step1: Solve for $S$ in $10S + 7=5\sqrt{3}+7$

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

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

We know that $\sin(\theta)=\frac{\sqrt{3}}{2}$ has solutions in the unit - circle. The principal solutions are $\theta=\frac{\pi}{3}+2k\pi$ and $\theta = \frac{2\pi}{3}+2k\pi$, where $k\in\mathbb{Z}$.

Step3: Solve for $\theta$ in $10\sin(\theta)+7 = 5\sqrt{3}+7$

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

Answer:

$S=\frac{\sqrt{3}}{2}$; $\theta=\frac{\pi}{3}+2k\pi,\frac{2\pi}{3}+2k\pi,k\in\mathbb{Z}$; $\theta=\frac{\pi}{3}+2k\pi,\frac{2\pi}{3}+2k\pi,k\in\mathbb{Z}$