Question 410944
Let {{{z = sin(B)}}}. Then, we have a quadratic


{{{z^2 - 3z + 2 = 0}}}. Factoring, this is equal to


{{{(z-2)(z-1) = 0}}} --> z = 2 or z = 1, sin(B) = 2 or sin(B) = 1 by back-substituting. However, sin(B) cannot be 2 since it is bounded by {{{-1 <= sin(B) <= 1}}}. Therefore, the only solutions occur when {{{sin(B) = 1}}} --> {{{B = pi/2}}}.