assignment 7.5: solving trigonometric equations\nscore: 0/100 answered: 0/10\nquestion 1\nfind all solutions…

assignment 7.5: solving trigonometric equations\nscore: 0/100 answered: 0/10\nquestion 1\nfind all solutions to ( 2 sin ( \theta ) = - sqrt { 3 } ) on the interval ( 0 leq \theta < 2 pi ).\n( \theta = )\ngive your answers as exact values in a list separated by commas.\nquestion help: > video 1 > video 2 d post to forum\nsubmit question
Answer
Explanation:
Step1: Solve for (\sin(\theta))
Divide both sides of (2\sin(\theta)=-\sqrt{3}) by (2). (\sin(\theta)=-\frac{\sqrt{3}}{2})
Step2: Find the reference angle
We know that (\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}), so the reference angle (\theta_{r}=\frac{\pi}{3})
Step3: Determine the quadrants
Since (\sin(\theta)<0), (\theta) is in the third or fourth quadrant. In the third quadrant: (\theta=\pi+\theta_{r}=\pi + \frac{\pi}{3}=\frac{4\pi}{3}) In the fourth quadrant: (\theta = 2\pi-\theta_{r}=2\pi-\frac{\pi}{3}=\frac{5\pi}{3})
Answer:
(\frac{4\pi}{3},\frac{5\pi}{3})