I'm not sure what I'm missing here:
Solve $\displaystyle a\sin(2\theta)-b\cos(2\theta)=\sin\theta$ for θ in terms of a and b.
Every attempt I've made has gotten rather messy.
--Kevin C.
I get to $\displaystyle 4(a^2+b^2)\cos^4\theta-4a\cos^3\theta+(1-4(a^2+b^2))\cos^2\theta+4a\cos\theta+(b^2-1)=0$, a quartic equation for $\displaystyle \cos\theta$. Attempting to solve that via the standard "double completing the square" method led to a horrible cubic.
--Kevin C.
I think a lot of times as math students we want a nice or "pretty" if you will solution to problems. Sometimes though, we get solutions that look messy and automatically think they must be wrong.
I do this all the time.
Eventually, you will learn to trust and not doubt yourself so much!