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.
Best Answer
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 $$\sin2x-\cos2x=0.$$