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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?

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    $\begingroup$ Try using $\sin2x=2\sin x\cos x$. $\endgroup$ – Ian Coley Feb 15 '14 at 20:40
  • $\begingroup$ @IanColey Nice approach, but there is the user who provided the hint already. $\endgroup$ – NasuSama Feb 15 '14 at 20:57
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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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  • $\begingroup$ Hehe. Changed back to before. $\endgroup$ – NasuSama Feb 15 '14 at 20:56
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Hint : write $9 = 9\sin^2(x)+9\cos^2(x)$

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HINT:

When you have only terms like $\sin^2x,\cos^2x,\sin x\cos x$

divide either sides by $\cos^2x,$

$$9\sec^2x+40\tan^2x=-42\tan x\iff9(1+\tan^2x)+40\tan^2x=-42\tan x$$

$$\iff49\tan^2x-42\tan x+9=0$$ which is a Quadratic Equation in $\tan x$

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