Whenever you divide both sides of an equation by something, you are assuming that the thing you're dividing by is nonzero, because dividing by 0 is not valid. So going from 2 \sin \theta \cos \theta = \sin \theta ...

The two issues are: When cancelling a factor, note that this is only possible when the factor is not zero; but the factor may be zero in the solution to the problem. In this case \sin\theta = 0 ...

You have \frac{1}{x^2+y^2} = 1+\left|\frac{2}{y^2/x^2+1}-1\right| = 1+\frac{|x^2-y^2|}{x^2+y^2}, or put differently 1 = x^2+y^2 + |x^2-y^2|. At this point solving by cases helps. Where |x|<|y| ...

As you noted, \cos(2\theta+\pi/2) = \operatorname{Re}(e^{i(2\theta+\pi/2)}). e^{i(2\theta+\pi/2)}=(e^{i\theta})^2e^{i\pi/2}=(\cos\theta+i\sin\theta)^2(i) =i(\cos^2\theta+2i\cos\theta\sin\theta-\sin^2\theta) ...

I have confused you with the definition of mean radius with the average ray length. When people think of average radius they think of "average squared minimized" or mean radius. What I am focusing on ...

You can choose C=c+\pi, and then \sin{(\theta+C)}=-\sin{(\theta+c)}, so having the \pm there doesn't create any more solutions if you allow any c \in [0,2\pi).

The right way to solve is: 2\cos^2{x}=2\sqrt{3}\cos{x}\sin{x} \to \cos x(\cos x-\sqrt{3}\sin x)=0 what give you: \cos x=0 \text{ or } \cos x-\sqrt{3}\sin x=0 For \cos x=0\to x=\frac{\pi}{2}+k\pi ...

Without using trigonometry: enjoy... By coloring triangles the same I am trying to emphasize the similar triangles, which are not necessarily equal (although red and blue triangles are). I don't mean ...