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This question is from a 2012 VMK entrance exam

I was trying to solve it first by expanding $\sin 4 x = 2 \sin 2 x \cos 2x$, then by noticing that if divided by 2, one can get, e.g. $ \frac{\sqrt{3}}{2} \cos 3 x = \sin \frac{\pi}{6} \cos 3 x$ and then trying to find some way to obtain $\sin (a +b)$ or $\cos (a+b)$, but without any success.

I suspect there should be some adding an extra term and then multiplying out, but can't get my head around it.

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4 Answers 4

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I tried writing $4x$ as $3x+x$ and $2x$ as $3x-x$ to get:$$\sin(3x+x)+\sqrt{3}\sin(3x)+\sin(3x-x)=0$$$$\therefore\sin(3x)\cos(x)+\cos(3x)\sin(x)+\sqrt{3}\sin(3x)+\sin(3x)\cos(x)-\cos(3x)\sin(x)=0$$$$\therefore2\sin(3x)\cos(x)+\sqrt{3}\sin(3x)=0$$$$\therefore\sin(3x)(2\cos(x)+\sqrt{3})=0$$Hopefully you can solve from here...

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Why not add $\sin 4x$ and $\sin 2x$? You'll get $2\sin 3x\cdot \cos x +\sqrt{3}\sin 3x = 0$...

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If you don't see any trick, you can always try \begin{align*} &\sin(4x)=\text{Im}(\cos x+i\sin x)^4=4\cos^3\!x\sin x-4\cos x\sin^3\!x\\ &\sin(2x)=\text{Im}(\cos x+i\sin x)^3=3\cos^2\!x\sin x-\sin^3\!x\\ &\sin(3x)=\text{Im}(\cos x+i\sin x)^2=2\cos x\sin x. \end{align*} Plugging it in yields $$(4\cos^3\!x-4\cos x\sin^2\!x+\sqrt3(3\cos^2\!x-\sin^2\!x)+2\cos x)\cdot\sin x=0$$ and you can solve the left factor by setting $y=\cos x$ \begin{align*} 0&=4y^3-4y(1-y^2)+\sqrt3(3y^2-(1-y^2))+2y=8y^3-4y+\sqrt3(4y^2-1)+2y\\ &=2y(4y^2-1)+\sqrt3(4y^2-1)=(2y+\sqrt3)(2y+1)(2y-1). \end{align*} I'm sure you could finish this.

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after applying the addition formulas we get this here $$\sin \left( x \right) \left( 8\, \left( \cos \left( x \right) \right) ^{3}+4\, \left( \cos \left( x \right) \right) ^{2}\sqrt {3}- 2\,\cos \left( x \right) -\sqrt {3} \right) =0$$

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