hello
I am thinking this should factorise as the difference of 2 squares or be converted using trig identities
I just keep hitting dead ends though
Any help would be appreciated
Hello, 200001!
This is not a difference of two squares . . .
$\displaystyle 3\sin^2\!\theta - \sin\theta\cos\theta - 4\cos^2\!\theta \:=\:0$
Factor: .$\displaystyle (\sin\theta + \cos\theta)(3\sin\theta - 4\cos\theta) \:=\:0$
And we have two equations to solve:
$\displaystyle \sin\theta \;+\; \cos\theta \:=\:0 \quad\Rightarrow\quad \sin\theta \:=\:-\cos\theta \quad\Rightarrow\quad \frac{\sin\theta}{\cos\theta} \:=\:-1 \quad\Rightarrow\quad \tan\theta \:=\:-1$
. . Hence: .$\displaystyle \theta \:=\:\arctan(-1)$
$\displaystyle 3\sin\theta \;-\; 4\cos\theta \:=\:0 \quad\Rightarrow\quad 3\sin\theta \:=\:4\cos\theta \quad\Rightarrow\quad \frac{\sin\theta}{\cos\theta} \:=\:\frac{4}{3} \quad\Rightarrow\quad \tan\theta \:=\:\frac{4}{3}$
. . Hence: .$\displaystyle \theta \:=\:\arctan\left(\tfrac{4}{3}\right)$
With fairly basic trig, you pretty much have less than 10 options total, and you can usually rule several of those out just by looking at what trig functions are being used. for example, in your problem you could have also divided by $\displaystyle sin^2$, but you would not have really wanted to use something like secant to tangent. You learn mostly with practice.