Find all values of $\theta$ such that the tangent line to $f(\theta) = \theta \sin(\theta)$ is given by $y = \theta$ $$f(\theta) = \theta \sin(\theta)$$
Derived to:
$$f'(\theta) = \sin(\theta) + \theta \cos(\theta)$$
How do I solve it from here?
 A: The generic tangent line to f at the point $(\theta_{0},f(\theta_{0}))$ is $$T(\theta)=f(\theta_{0})+f'(\theta_{0})(\theta-\theta_{0})=\theta_{0}\cdot \sin(\theta_{0})+(\sin(\theta_{0})+\theta_{0}\cos(\theta_{0}))(\theta-\theta_{0})$$
You need $\theta_{0}$ such that $f(\theta_{0})-f'(\theta_{0})\theta_{0}=0$ and $f'(\theta_{0})=1$; that is, $f'(\theta_{0})=1$ and $f(\theta_{0})=\theta_{0}$.  So, you need to solve: $$\sin(\theta_{0})+\theta_{0}\cos(\theta_{0})=1$$ and $$\theta_{0}\sin(\theta_{0})=\theta_{0}$$ for $\theta$.  To satisfy the second equation, you clearly need $\theta=\frac{\pi}{2}+2n\pi$ for any integer $n$, since this is the set of angles which have a sine value of $1$.  If you plug this set of angles into the first equation, you see that since each has a cosine value of $0$, the first equation automatically holds.  Thus, your answer is the set $\{\frac{\pi}{2}+2n\pi|n\in \mathbf{Z}\}$... or just $\frac{\pi}{2}$ if you are restricting your answer to angles between $0$ and $2\pi$. 
