1. ## Rewrite this identity as a quadratic equation in tan(x)

"Show that the equation $\displaystyle sin^2x + 3 sin\ x\ cos\ x = 2$ can be written as a quadratic equation in $\displaystyle tan\ x$."

I can't see any way to get a quadratic equation with no other trig functions than $\displaystyle tan\ x$. I tried dividing throughout by $\displaystyle cos\ x$ etc. to generate $\displaystyle tan\ x$, but always ended up with a leftover sin or cos somewhere.

Could anyone give me some hints?

2. Originally Posted by EvilKitty
"Show that the equation $\displaystyle sin^2x + 3 sin\ x\ cos\ x = 2$ can be written as a quadratic equation in $\displaystyle tan\ x$."

I can't see any way to get a quadratic equation with no other trig functions than $\displaystyle tan\ x$. I tried dividing throughout by $\displaystyle cos\ x$ etc. to generate $\displaystyle tan\ x$, but always ended up with a leftover sin or cos somewhere.

Could anyone give me some hints?
Divide the whole equation by $\displaystyle \cos^2 x$, and remember that $\displaystyle \frac{1}{\cos^2 x}=1+\tan^2 x$.
Note that you don't lose anything by doing that division: an x for which $\displaystyle \cos x=0$ could not possibly be a solution of the original equation.

3. It worked, thanks! I got $\displaystyle -tan^2\ x\ + 3tan\ x -2 = 0$.

How did you get $\displaystyle \frac{1}{\cos^2 x}=1+\tan^2 x$? I can't see how to derive this from the trigonometric identities I've learnt so far.

4. Originally Posted by EvilKitty
It worked, thanks! I got $\displaystyle -tan^2\ x\ + 3tan\ x -2 = 0$.

How did you get $\displaystyle \frac{1}{\cos^2 x}=1+\tan^2 x$? I can't see how to derive this from the trigonometric identities I've learnt so far.
Just use the fact that $\displaystyle \cos^2x+\sin^2x = 1$, for all x, to replace the numerator by something a little more interesting:

$\displaystyle \frac{1}{\cos^2x}=\frac{\cos^2 x+\sin^2 x}{\cos^2 x}=\frac{\cos^2 x}{\cos^2 x}+\frac{\sin^2 x}{\cos^2 x}=1+\tan^2 x$