<s>[INST] 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
 [/INST] Simple factorization..
$$2 \cos^2 x + 5 \cos x + 2 =0$$
$$(2 \cos^2 x + 4 \cos x) +(\cos x + 2) =0$$
$$2 \cos x(\cos x +2)+( \cos x + 2) =0$$
$$(\cos x +2)( 2\cos x + 1) =0$$
$$\implies ( 2\cos x + 1) =0$$
$$\implies \cos x =-\frac 12$$
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