<s>[INST] How can this trig equation be simplified? We have $9+40\sin^2x=-42\sin x\cos x$.
I know this simplifies to $7\sin x+3\cos x=0$, but how?
 [/INST] Along with Trafalgar Law's hint, we can use the fact that $\sin^2(x) + \cos^2(x) = 1$.  Multiply both sides by $9$ to get $9 = 9\sin^2(x) + 9\cos^2(x)$.  We then have
$$\begin{aligned}
9\sin^2(x) + 9\cos^2(x) + 40\sin^2(x) &= -42\sin(x)\cos(x)\\
49\sin^2(x) + 42\sin(x)\cos(x) + 9\cos^2(x) &= 0\\
(7\sin(x) + 3\cos(x))^2 &= 0
\end{aligned}$$
which gives the desirable equation you want to have.
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