calculus
Find the points on the curve where the tangent is horizontal or vertical.
$x = \cos(3 \theta)$ , $y = 5 \sin(\theta)$
I have done quite a few of these problems and this was the first one that I keep getting wrong and coming back with same answer.
Horizontal tangents: $(0,-5) , (0,5)$
Vertical tangent $(-1,0), (1,0)$
Any ideas?
$$\dfrac{dy}{d\theta}\cdot \frac{d\theta}{dx} = \dfrac{dy}{dx}$$ and we want to know when $\dfrac{dy}{dx} = 0$.
$$\frac{dx}{d\theta} = \frac{\cos3\theta}{r},\quad \frac{dy}{d\theta} = \frac{\sin 3\theta}{r} \implies \dfrac{dy}{dx} = \frac{\sin{3\theta}}{r}\cdot \dfrac{r}{\cos 3\theta}$$
Now solve for when $$\require{cancel} \dfrac{dy}{dx} = \frac{\sin{3\theta}}{\cancel{r}}\cdot \dfrac{\cancel{r}}{\cos 3\theta} = = \tan(3\theta)= 0$$
That's going to happen when $\sin 3\theta = 0$ (and note that when $\sin{3\theta} = 0 \iff 3\theta = k \pi \iff \theta = k\pi/3\;$ ($k$ any integer), we have that $\cos 3\theta = \pm 1\iff 3\theta = k\pi \iff \theta = k\pi/3$.)
The points at which the tangents are horizontal and/or vertical are given in polar coordinates, $(r, \theta)$.
You have calculated $$ \frac{dx}{d\theta}=-\sin\theta(1+4\cos\theta) $$ and $$ \frac{dy}{d\theta}=\cos\theta+2\cos2\theta. $$ Since $\cos2\theta=2\cos^2\theta-1$, the $\frac{dy}{d\theta}$ simplifies to $$ \frac{dy}{d\theta}=\cos\theta+4\cos^2\theta-2. $$
Vertical and horizontal tangents
In fact, we do not need to find the values of $\theta$ if we want to find the points where we have horizontal/vertical tangents. It is enough to know $\cos\theta$ and $\sin\theta$ and then use the formulas $$ x=r\cos\theta=(1+2\cos\theta)\cos\theta,\quad\text{and}\quad y=r\sin\theta=(1+2\cos\theta)\sin\theta. $$ Anyways, I indicate how to solve the equations. First, we can assume that $0\leq \theta<2\pi$, since $r=1+2\cos\theta$ is $2\pi$ periodic.
Now, one should solve $\frac{dx}{d\theta}=0$ to find the vertical tangent lines. For the vertical ones, $dx/d\theta=0$. Since the product $\sin\theta(1+4\cos\theta)=0$ if and only if (at least) one of its factors is zero, we find that either $\sin\theta=0$ (this gives $\theta=0$ and $\theta=\pi$), or $\cos\theta=-1/4$ (this gives $\theta=\arccos(-1/4)$ or $\theta=2\pi-\arccos(-1/4)$. In total, we get four values of $\theta$.
To find the horizontal tangent lines we solve $dy/d\theta=0$. We note that $dy/d\theta$ is a second degree polynomial in $\cos\theta$. Thus, with $t=\cos\theta$, we should solve $4t^2+t-2=0$. This has solutions $t=\frac{1}{8}(-1\pm\sqrt{33})$. So, we find $\theta$ by solving $$ \cos\theta=\frac{1}{8}(-1+\sqrt{33})\approx 0.59,\quad\text{and},\quad \cos\theta=\frac{1}{8}(-1-\sqrt{33})\approx -0.84. $$ To calculate $\sin\theta$ you can use the fact that $\cos^2\theta+\sin^2\theta=1$. I leave it to you to fill in the rest of the details. If you succeed, you will probably find something like the picture below, where I have drawn the curve $r=1+2\cos\theta$ together with the points I got by doing these calculations (red for horizontal tangents, green for vertical ones).
Best Answer
In the $xy$ Cartesian coordinate system, horizontal tangents occur when $\frac{dy}{dx} = 0$ and vertical tangents occur when $\frac{dx}{dy} = 0$. Since $x$ and $y$ are given separately as functions of $\theta$, it'll likely be simplest and easiest to determine these derivatives implicitly by using, for $\frac{dy}{dx}$, the fraction of $\frac{dy}{d\theta}$ divided by $\frac{dx}{d\theta}$, with $\frac{dx}{dy}$ being the reciprocal. We thus get that
$$\frac{dy}{d\theta} = \frac{d\left(5\sin\left(\theta\right)\right)}{d\theta} = 5\cos\left(\theta\right) \tag{1}\label{eq1}$$ $$\frac{dx}{d\theta} = \frac{d\left(\cos\left(3\theta\right)\right)}{d\theta} = -3\sin\left(3\theta\right) \tag{2}\label{eq2}$$
Thus, for example,
$$\frac{dy}{dx} = \frac{5\cos\left(\theta\right)}{-3\sin\left(3\theta\right)} \tag{3}\label{eq3}$$
so $\frac{dx}{dy} = \frac{-3\sin\left(3\theta\right)}{5\cos\left(\theta\right)}$. Because you're looking for values where the derivative is $0$, you only need to check the numerator being $0$ and ensure the denominator is not $0$. However, note you need to consider all valid values where the numerator is $0$, i.e., using that the $0$ values for $\sin$ and $\cos$ are periodic with a period of $\pi$ radians. After doing this, you should have more results than what you've indicated. I trust you can finish the rest yourself.