Let's check the given function:
[tex]\sin \theta=\frac{\sqrt[]{3}}{2}[/tex]
Where sinθ is positive in quadrants 1 and 2.
Also, sinθ= opposite side/ hypotenuse:
Then:
Now, the first angle can be found solving theta:
[tex]\theta=\arcsin (\frac{\sqrt[]{3}}{2})[/tex][tex]\theta=60\text{ or }\theta=\frac{\pi}{3}[/tex]
To find the second angle we use:
[tex]\theta_2=\pi-\theta[/tex]
Then:
[tex]\theta_2=\pi-\frac{\pi}{3}[/tex][tex]\theta_2=\frac{2}{3}\pi\text{ or }\theta_2=120[/tex]
Now, sinθ will be again positive when we complete a whole circle.
Then, we use
If a circle has 2π, then 2π+1/2π =5/2π is when sinθ is positive again.
There, we use:
120* 4= 480
Then:
[tex]\sin (480)=\frac{\sqrt[]{3}}{2}[/tex]
Hence:
[tex]\theta_3=\frac{8}{3}\pi\text{ or }\theta_3=480[/tex]
