Find the equation of the left most tangent to the curve $$r=3+2\cos(\theta)$$
I tried using the formula of the slope of the tangent to the polar curve given by: $$\frac{d y}{d x}=\frac{\frac{d r}{d \theta} \sin \theta+r \cos \theta}{\frac{d r}{d \theta} \cos \theta-r \sin \theta}$$
For vertical tangency we have: $$\frac{dr}{d\theta}\cos(\theta)=r\sin(\theta)$$ Now $\frac{dr}{d\theta}=-2\sin(\theta)$
So we get $$3\sin(\theta)=-4\cos(\theta)\sin(\theta)$$ $\implies$ $\sin(\theta)=0$ or $\cos(\theta)=\frac{-3}{4}$ Now how to find the equation of left most vertical tangent?
The most straightforward method is to observe that the leftmost vertical tangent corresponds to a value of $\theta$ such that $x(\theta)$ is a minimum. That is to say, $$x(\theta) = (3 + 2 \cos \theta) \cos \theta$$ is minimized. This corresponds to solving $$0 = \frac{d x}{d \theta} = -3 \sin \theta - 4 \cos \theta \sin \theta = - \sin \theta (3 + 4 \cos \theta).$$ Hence the candidates are $\theta \in \{0, \pi, \arccos (-\frac{3}{4}), 2\pi - \arccos (-\frac{3}{4}) \}$. Substitution into $x(\theta)$ eliminates the $0$ and $\pi$, and the other two angles both yield $x = -9/8$. This is the required equation of the tangent line.