1. find all possible angles for θ using the domain ( 0 leq \theta < 2pi ).\na) ( sin \theta =…

1. find all possible angles for θ using the domain ( 0 leq \theta < 2pi ).\na) ( sin \theta = \frac{sqrt{3}}{2} )\nb) ( cos \theta = -\frac{1}{2} )\nc) ( cos \theta = 1 )\nd) ( cos \theta = \frac{sqrt{2}}{2} )

1. find all possible angles for θ using the domain ( 0 leq \theta < 2pi ).\na) ( sin \theta = \frac{sqrt{3}}{2} )\nb) ( cos \theta = -\frac{1}{2} )\nc) ( cos \theta = 1 )\nd) ( cos \theta = \frac{sqrt{2}}{2} )

Answer

Explanation:

Step1: Recall the unit - circle values

For (y = \sin\theta), (\sin\theta=\frac{\sqrt{3}}{2}). The reference angle (\theta_{r}=\frac{\pi}{3}). Since (\sin\theta>0), (\theta) is in the first and second quadrants. (\theta=\frac{\pi}{3}) (first - quadrant) and (\theta=\pi-\frac{\pi}{3}=\frac{2\pi}{3}) (second - quadrant)

Step2: For (x = \cos\theta), (\cos\theta=-\frac{1}{2})

The reference angle (\theta_{r}=\frac{\pi}{3}). Since (\cos\theta<0), (\theta) is in the second and third quadrants. (\theta=\pi-\frac{\pi}{3}=\frac{2\pi}{3}) (second - quadrant) and (\theta=\pi+\frac{\pi}{3}=\frac{4\pi}{3}) (third - quadrant)

Step3: For (\cos\theta = 1)

We know that (\cos\theta=x) on the unit circle. When (x = 1), (\theta = 0) (since (\cos(0)=1) and (0\leq\theta<2\pi))

Step4: For (\cos\theta=\frac{\sqrt{2}}{2})

The reference angle (\theta_{r}=\frac{\pi}{4}). Since (\cos\theta>0), (\theta) is in the first and fourth quadrants. (\theta=\frac{\pi}{4}) (first - quadrant) and (\theta = 2\pi-\frac{\pi}{4}=\frac{7\pi}{4}) (fourth - quadrant)

Answer:

a) (\theta=\frac{\pi}{3},\frac{2\pi}{3}) b) (\theta=\frac{2\pi}{3},\frac{4\pi}{3}) c) (\theta = 0) d) (\theta=\frac{\pi}{4},\frac{7\pi}{4})