Trying to understand an inverse trig function.
Evaluate Cos^-1 (sqr3/2)
So I know it's 30 degrees, but wouldn't it also be 150 degrees because the Arc-cos is defined in the 2nd quadrant?
Think of it as the angle Ø so that
cos Ø = √3/2
recall that the cosine is + in quads I and IV
so you are right, one solution is 30°
but the other must be 360-30 = 330 (the IV quad angle)
Actually there is an infinite number of solutions,
others are -30, -330 , 390 , etc
Your choice of 150 would be incorrect,
cos 150 = -√/2