how to solve $\frac{\sin^2(\theta)-3\cos^2(\theta)+1}{\sin^2(\theta)-\cos^2(\theta)}\equiv2$ and similar equations

Question

I was wondering how to solve an equation where there is a fraction with $\sin^2(\theta)$ and $\cos^2(\theta)$ on top and bottom. The specific equation i am having a problem with is:

$\frac{\sin^2(\theta)-3\cos^2(\theta)+1}{\sin^2(\theta)-\cos^2(\theta)}\equiv2$

My original idea was to use $\sin^2(\theta)+\cos^2(\theta)\equiv1$ and turn $\frac{\sin^2(\theta)-3\cos^2(\theta)+1}{\sin^2(\theta)-\cos^2(\theta)}\equiv2$ into $\frac{4\sin^2(\theta)-2}{2\sin^2(\theta)-1}\equiv2$ but that does not work as you just get to $1\equiv1$.

Then I tried turning it into $\frac{4\sin^2(\theta)-2}{1-2\cos^2(\theta)}\equiv2$ but then I get stuck at $\frac{4\sin^2(\theta)}{4-2\cos^2(\theta)}\equiv1$

I have the thought that you will have to use $\frac{\sin(\theta)}{\cos(\theta)} = \tan(\theta)$ but I dont know how to get there.

In general how do you solve equations like these. Thanks very much for any help.

Answer 1

You can proceed and solve question step by step using Trignometric identities.

Let's take left hand side: $\frac{\sin^2(\theta)-3\cos^2(\theta)+1}{\sin^2(\theta)-\cos^2(\theta)}$

$=\frac{\sin^2(\theta)-3\cos^2(\theta)+\sin^2(\theta)+\cos^2(\theta)}{\sin^2(\theta)-\cos^2(\theta)}$ (Since ${\sin^2(\theta)+\cos^2(\theta)}$=1)

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

$=2=$Right hand side

Note: Things are not complicate. We tend to make it. ;-)

Answer 2

What you did so far is fine.

You may conclude that any $\theta$ is a solution except those that satisfy $\sin^2\theta =\frac12$, which prevents the denominator to be zero.

Answer 3

Simply use the duplication formulæ: $$\sin^2\theta-\cos^2\theta=-\cos 2\theta,$$ $$\sin^2(\theta)-3\cos^2(\theta)+1=2-4\cos^2\theta=-2\cos 2\theta.$$