If sin theta = (sqrt 3)/2, which could not be the value of theta? A. 60 degrees B. 120 degrees C. 240 degrees D. 420 degrees I know that the solutions for theta are pi/3 and 2pi/3, but how does that correlate?
Your are right, [tex] \pi /3[/tex] and 2[tex] \pi /3[/tex] are solutions. Of course each of these solutions can be replaced by the same number +k2[tex] \pi [/tex] (with k being an element of Z) since sin x=sin (x+k2[tex] \pi [/tex]) with k being an element of Z. Those are angle measures expressed in radians. If you translate in degrees you basically have to know that [tex] \pi [/tex] radian=180 degrees so [tex] \pi /3[/tex] radian=60 degrees and 2[tex] \pi /3[/tex]=120 degrees. So this two could be values of theta (A and B). On top of that [tex] \pi /3+2 \pi [/tex] could also be a solution (x+k2[tex] \pi [/tex] with k=1). This can be translated to 60+360=420 degrees which is solution D. So C. 240 degrees is the only one that could not be a value of theta.
