Find the number of solutions to $\sin x+\sin2x+\sin3x=\cos x+\cos2x+\cos3x$ with $x\in[0,2\pi)$

I tried to solve it by doing this:-

$$2\sin(2x)\cos x + \sin(2x) = 2\cos(2x)\cos x + \cos(2x)$$

$$\sin(2x) (2\cos x+1) = \cos(2x) (2\cos x + 1)$$

$$ \sin(2x) = \cos(2x)$$

$$ 2\cos x\sin x = 1 - 2\sin^2 x$$

$$ 2\cos x\sin x + 2\sin^2 x - 1 = 0$$

$$ 2\sin x(\cos x + \sin x) -1=0$$

$$ 2\sin x(2\sin x) -1=0$$

$$ 4\sin^2 x = 0$$

$$ \sin^2 x = (\frac{1}{2})^2$$

$$ \sin^2 x = \sin^2 (\frac{\pi}{6})$$

Now by using $x = nπ -\frac{\pi}{6}$ I got 4 solutions between $0,2\pi$. But the answer is 6. Where I did wrong. Please don't mind for mistakes and answer ASAP. THANKS IN ADVANCE.

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    $\begingroup$ Welcome to MSE. Please use MathJax to format your posts. You'll get a lot more help if your questions are easy to read. $\endgroup$ – saulspatz May 28 at 14:11
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    $\begingroup$ Also I think it's impolite to say 'answer ASAP'. Kindly note that this site objective is not merely to just answer your question. $\endgroup$ – Mann May 28 at 14:15

You have also $$2\cos{x}+1=0,$$ which gives another two roots.

Because $$\sin2x(2\cos{x}+1)=\cos2x(2\cos{x}+1)$$ it's $$\sin2x(2\cos{x}+1)-\cos2x(2\cos{x}+1)=0$$ or $$(2\cos{x}+1)(\sin2x-\cos2x)=0$$ and we obtain: $$2\cos{x}+1=0$$ or $$\sin2x-\cos2x=0.$$

  • $\begingroup$ Explain please.. $\endgroup$ – Diya Sarkar May 28 at 14:20
  • $\begingroup$ @DiyaSarkar You divided out $2\cos x+1$ at one point, but you can't do that if $2\cos x+1=0$, so you have to check those roots too. $\endgroup$ – J.G. May 28 at 14:22
  • $\begingroup$ So what should I do there??? $\endgroup$ – Diya Sarkar May 28 at 14:22
  • $\begingroup$ @Diya Sarkar I added something. See now. $\endgroup$ – Michael Rozenberg May 28 at 14:23

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