Trigonometry and quadratics : Possible mismatch? There’s this problem I came across, gives me an invalid answer by using general quadratic formula. Wonder why?
$2\sin^2{x} -5\cos{x} -4 =0 $
Here’s what I did:
$2\sin^2{x} -5\cos{x} -4 =0 $
$2(1-\cos^2{x}) - 5 \cos{x} - 4 = 0$
$2 \cos^2{x} + 5 \cos{x} + 2 =0$
This is a quadratic function of $\cos x$, thus,
$\cos{x} = (-5 +  3)/4$ or $\cos{x}= (-5 - 3)/4$
But, the answer given is $\cos{x}=-\frac{1}{2} $ and WolframAlpha says the same but doesn’t show how to do it.
What did I do wrong?
Update: Thank you very much, everyone. Turns out that I wrote the squareroot of 9 as squareroot of 3. My bad

Yes from here (you are right) we have:
$$2\sin^2x -5\cos x -4 =0 \implies 2\cos^2 x+5\cos x +2=0$$
that is by quadratic formula (here was your mistake):
$$\cos x = \frac{-5\pm \sqrt{25-16}}{4}= \frac{-5\pm 3}{4}\implies \cos x=-\frac12$$
