Solving $\frac{\cos^23t}{\tan t}+\frac{\cos^2t}{\tan3t}=0$

Question

I am having trouble solving the trigonometric equation below: $$\frac{\cos^2\left(3t\right)}{\tan\left(t\right)}+\frac{\cos^2\left(t\right)}{\tan\left(3t\right)}=0$$

I would appreciate any help you can give me! Thank you!

Answer 1

I will do the first one, hopefully, it will give you some ideas how to tackle the rest.

$\frac {\cos^2 3t}{\tan t} + \frac {\cos^2 t}{\tan 3t} = 0$

Find the common denominator and make this into one fraction.

$\frac {\cos^2 3t\tan 3t + \cos^2t\tan t}{\tan 3t\tan t} = 0$

Lets focus on the numerator. If this expression equals $0$ then the numerator equals $0$ and the denominator does not equal $0.$

$\cos^2 3t\tan 3t + \cos^2t\tan t = 0$

use: $\tan t = \frac {\sin t}{\cos t}$ and cancel.

$\cos 3t\sin 3t + \cos t\tan t = 0$

I want to use $2\cos t\sin t = \sin 2t$

$2\cos 3t\sin 3t + 2\cos t\tan t = 0\\ \sin 6t + \cos t\tan t = 0$

The next step is a little bit obscure... $\sin 3t = (4\sin t - 3\sin^3 t)$

$3\sin 2t - 4\sin^3 2t + \sin 2t = 0$

Everything is in terms of $\sin 2t$!

$u=\sin 2t$ will make what we have look more like a polynomial.

$4u - 4u^3 = 0\\ u(1-u)(1+u)= 0\\ u = 0, \pm 1$

Reverse the substitution.

$\sin 2t = 0 \text { or } \pm 1$

$t = 0, \pm\frac {\pi}{2}, \pm \frac {\pi}{4}, \pm \frac{3\pi}{4}+ n\pi$

But if $t=0$ the denominator in the original expression equals 0, and we are going to throw that answer away. And $\tan t$ is not defined at $t = \pm\frac {\pi}{2}$ so we should disregard that solution, too.

$t = \frac {(2n+1)\pi}{4}$

Answer 2

Here it is a slightly different approach from @Doug M. One has that \begin{align*} \frac{\cos^{2}(3t)}{\tan(t)} + \frac{\cos^{2}(t)}{\tan(3t)} = 0 & \Rightarrow \cos^{2}(3t)\tan(3t) + \cos^{2}(t)\tan(t) = 0\\\\ & \Rightarrow \cos(3t)\sin(3t) + \cos(t)\sin(t) = 0\\\\ & \Rightarrow \sin(6t) + \sin(2t) = 0\\\\ & \Rightarrow \sin(6t) = \sin(-2t) \end{align*}

After solving such equation, verify whether the solutions satisfy $\tan(t)\tan(3t)\neq 0$.

Can you take it from here?