First thing to realise is that it can't be as evil as it looks! First, get rid of that fraction and bring the left hand denominator over to the right, which becomes 2cos a(cos 5a + 5cos 3a + 10cos a). My first thought was to reduce everything to cos a, but then I thought, no, that's going to be horrible. Then I started thinking about expanding cos(A+B) (= cos A cos B - sin A sin B) and cos(A-B) (= cos A cos B + sin A sin B). Add them together and the sines drop out: 2cos A cos B. That's a bit like our problem, because there are no sines. For example, cos 6a = cos(5a+a) = cos 5a cos a - sin 5a cos a and cos 4a = cos(5a-a) = cos 5a cos a + sin 5a sin a. Add them together and we get 2cos 5a cos a. Now rewrite the left hand side: (cos 6a + cos 4a) + (5cos 4a + 5cos 2a) + 10cos 2a + 10. That gives us 2cos 5a cos a + 10cos 3a cos a + 10cos 2a + 10. On the right hand side we have, after opening the brackets, 2cos5a cos a + 10 cos 3a cos a + 20 cos^2 a. Miraculously, two terms cancel out! That leaves 10cos 2a + 10 = 20 cos^2 a. Expand cos 2a and we get 2cos^2 a - 1. So, 20cos^2 a = 20cos^ a. That means the original equation was not an equation to be solved, it's an identity - true for all values of a. We can show this by substituting a in the original equation, using for example a=0. We get (1+6+15+10)/(1+5+10) = 2. When we try to substitute a = 90, we end up dividing zero by zero, with zero as the result. Put a = 60 and we get 0.5/0.5 = 1.