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Solving the equation: $$3\cos x - \sin 2x = \sqrt{3}(\cos 2x + \sin x)\tag{1}$$

I tried to write $(1)$ becomes $$\sqrt{3}\sin \left(\frac{\pi}{3}-x\right)=\sin \left(\frac{\pi}{3}+2x \right)$$

Now, I have stuck :( , can you help me?

Thanks!

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After staring at this for a while I think I've got it. Assuming what you've done so far is correct

$$\sin(\frac\pi3+2x)=\sin[\pi-(\frac\pi3+2x)]=\sin(\frac{2\pi}3-2x)$$

Substituting this gives

$$\sqrt3\sin(\frac\pi3-x)=\sin(\frac{2\pi}3-2x)$$ $$\sqrt3\sin(\frac\pi3-x)=2\sin(\frac\pi3-x)\cos(\frac\pi3-x)$$ $$\sin(\frac\pi3-x)[\sqrt3-2\cos(\frac\pi3-x)]=0$$

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  • $\begingroup$ Yes, it's what I need. Thank you very much. Anyway, I'm sure that I'm right. :) $\endgroup$ – kimtahe6 Dec 11 '14 at 9:46
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    $\begingroup$ Here's an other solution :): $$(1) \iff \sqrt{3}\cos(x+\frac{\pi}{6})=\sin(2x+\frac{\pi}{3}) \\ \iff \sqrt{3}\cos(x+\frac{\pi}{6})=2\sin(x+\frac{\pi}{6})\cos(x+\frac{\pi}{6})$$ $\endgroup$ – kimtahe6 Dec 12 '14 at 14:50
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Hint $$\cos{x}(3-2\sin{x})=\sqrt{3}(2\sin{x}+1)(1-\sin{x})$$ $$\Longrightarrow \left(\cos{\dfrac{x}{2}}-\sin{\dfrac{x}{2}}\right)\left(\cos{\dfrac{x}{2}}+\sin{\dfrac{x}{2}}\right)(3-2\sin{x})=\sqrt{3}\left(\cos{\dfrac{x}{2}}-\sin{\dfrac{x}{2}}\right)^2(2\sin{x}+1)$$ (1):$$\left(\cos{\dfrac{x}{2}}-\sin{\dfrac{x}{2}}\right)=0\Longrightarrow x=\dfrac{\pi}{2}+2k\pi$$ (2):$$\left(\cos{\dfrac{x}{2}}+\sin{\dfrac{x}{2}}\right)(3-2\sin{x})=\sqrt{3}\left(\cos{\dfrac{x}{2}}-\sin{\dfrac{x}{2}}\right)(2\sin{x}+1)\tag{1}$$ let $$\sin{\dfrac{x}{2}}+\cos{\dfrac{x}{2}}=t\Longrightarrow \sin{x}=t^2-1,\left(\sin{\dfrac{x}{2}}-\cos{\dfrac{x}{2}}\right)^2=2-t^2$$ use $(1)^2$ then we have $$t^2\cdot(3-2t^2+2)^2=3(2-t^2)(2t^2-1)^2\Longrightarrow 16t^6-56t^4+52t^2-6=0$$ $$\Longrightarrow (2t^2-3)(2t^2-2t-1)(2t^2+2t-1)=0$$ so $$\Longrightarrow t=\sqrt{\dfrac{3}{2}},-\sqrt{\dfrac{3}{2}},\dfrac{1}{2}(-1\pm \sqrt{3})-\dfrac{1}{2}(1+\sqrt{3})$$

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  • $\begingroup$ I know your hint when I use wolframalpha.com, but can you solve:$$ \left(\cos{\dfrac{x}{2}}+\sin{\dfrac{x}{2}}\right)(3-2\sin{x})=\sqrt{3} \left(\cos{\dfrac{x}{2}}-\sin{\dfrac{x}{2}}\right)(2 \sin{x}+1)$$????? math110 $\endgroup$ – kimtahe6 Dec 11 '14 at 6:31

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