cos(t) + sin(2t) = 0?

Question

Find the exact solutions of the equation that are in the interval [0, 2*Pi):

Answer 1

cos(t) + 2sin(t)cos(t) = 0, using double-angle identity for sin
cos(t)[1 + 2 sin(t)] = 0
so either cos(t) = 0, which means t=Pi/2 or 3Pi/2
(using fact that cos(-t)=cos(t), so cos(-Pi/2)=0, then using cos(t+2Pi)=cos(t), giving 3Pi/2)
or sin(t) = -1/2, which means t=11Pi/6 or 7Pi/6
(using sin(Pi/6)=1/2, and sin(-t)=-sin(t), so sin(-Pi/6)=-1/2, but sin(t+2Pi)=sin(t), so sin(11Pi/6)=-1/2. also, sin(Pi-t)=sin(t), so sin(5Pi/6)=1/2, so sin(-5Pi/6)=-1/2, add 2Pi and get sin(7Pi/6)=-1.2. Symmetries of sin and cos are essential to find all solutions!)