This stuff is throwing me for a loop! http://imgur.com/CcXfKvh
Anyway the problem involves trigonometric identities.

Question

anonymous · Answer

a¡in 1º equation after performing operations you get:
$sin \theta cos\theta +sin \theta(2-sin\theta)=0$
so one solution will be when either $sin \theta=0$, which happens at $\theta=pi/2, 3\pi/2$. The other could be when $cos\theta=sin \theta -2$ or $sin\theta-cos\theta =2$. But this is imposible. So the only solutions are $\theta=pi/2, 3\pi/2$.

anonymous · Answer

other 2 are strait forward

anonymous · Answer

sry , $sin\theta=0$ at $\theta=0 $ and $\theta =\pi$. I don't know why, but I was thinking of $cos\theta$