How does $e^{2\theta}(2\cos{\theta}-\sin{\theta})=0$ imply $\tan{\theta}=2$?

Question

The solution to a problem in my textbook seems to make up a trig identity out of nowhere. Here's the question:

Find the points on the spiral $r=e^{2\theta}, 0\leq\theta\leq\pi$, where the tangents are

a) perpendicular to the initial line

b) parallel to the initial line.

Give your answers to three significant figures.

The given solution for part a is:

$x=r\cos{\theta}=e^{2\theta}\cos{\theta}$

$\frac{dx}{d\theta}=0\implies 0=2e^{2\theta}\cos{\theta}-e^{2\theta}\sin{\theta}$

$0=e^{2\theta}(2\cos{\theta}-\sin{\theta})$

$\implies\tan{\theta}=2$

etc.

How did all that turn into a tan? What identity am I missing?

Answer 1

The exponential function can never be $0$. Since it is multiplied with something and $0$ is obtained, the other part must be zero: $$2\cos\theta-\sin\theta=0\tag1$$ This rearranges into $$2\cos\theta=\sin\theta\implies\frac{\sin\theta}{\cos\theta}=\tan\theta=2$$ Even by my standards I agree that writing $(1)$ as an intermediate step would be clearer, because the "no zero divisors" property of the real numbers is not commonly seen in trigonometry.

Answer 2

1. In general, \begin{align}&\sin\theta=k\cos\theta\\\iff{}& \tan\theta=k.\end{align}

   Proof

   1. Observe that if the LHS is true, then $\cos\theta\ne0,$ otherwise $\sin\theta=0=\cos\theta,$ which is impossible. Dividing the LHS statement by $\cos\theta$ gives the RHS statement.

   2. Multiplying the RHS statement by $\cos\theta$ gives the LHS statement.

2. $$0=e^{2\theta}(2\cos{\theta}-\sin{\theta})\iff0=2\cos{\theta}-\sin{\theta}\iff\tan\theta=2;$$ the first equivalence is valid because for each real $\theta,\;e^{2\theta}>0.$