# Solving $\cos(t\cos\theta)\cos(t\sin\theta) \sin 2\theta = \sin(t\cos\theta)\sin(t\sin\theta)$

I am trying to find all $$\theta\in\mathbb{R}$$ such that $$\cos(t\cos\theta)\cos(t\sin\theta) \sin 2\theta = \sin(t\cos\theta)\sin(t\sin\theta)$$ holds for all $$t\in\mathbb{R}$$. Just by looking at the equation it is clear that $$\theta = \frac{\pi n}{2}, n \in\mathbb{Z},$$ are solutions.

I think that there are no more solutions, since if we assume that $$\theta \not= \frac{\pi n}{2},$$ we can rearrange the equation to get $$\sin 2\theta = \frac{\sin(t\cos\theta)\sin(t\sin\theta)}{\cos(t\cos\theta)\cos(t\sin\theta)} = \tan(t\cos\theta)\tan(t\sin\theta).$$ Is it safe to conclude from here that the RHS will in general depend on $$t$$, so no single choice of $$\theta$$ will work for all $$t$$? Is there a better way of seeing this?