<s>[INST] Solving $1+\sin(x)+\sin(2x)+\sin(3x)=\cos(x)+\cos(2x)+\cos(3x)$ I'm trying to solve the following trigonometric equation

$$1+\sin(x)+\sin(2x)+\sin(3x)=\cos(x)+\cos(2x)+\cos(3x)$$

but I can't get any further than
$\iff 1 + 2\sin(2x)\cos(x) + \sin(2x) = 2\cos(2x)\cos(x) + \cos(2x)$
$\iff 1 + \sin(2x)(2\cos(x) + 1) = \cos(2x)(2\cos(x) + 1)$
$\iff 1 + \sin(2x)(2\cos(x) + 1) = \cos(2x)(2\cos(x) + 1)$
$\iff 1 = (\cos(2x)-\sin(2x))(2\cos(x) + 1)$
Could someone point me out in the right direction?
 [/INST] Let us use the Swiss army knife of trigonometry by setting $t = \tan(\frac{x}{2})$. One gets 
\begin{align}
\sin{x} &= \frac{2t}{1+t^2} \\
\sin{2x} &= 2\sin{x}\cos{x} = \frac{4t(1-t^2)}{(1+t^2)^2}\\
\sin{3x} &= 3\sin{x}- 4\sin^3{x} =  \frac{6t}{1+t^2} - \frac{32t^3}{(1+t^2)^3} \\
\cos{x} &= \frac{1-t^2}{1+t^2} \\
\cos{2x} &= \cos^2{x} - \sin^2{x} = \frac{t^4-6t^2 + 1}{(1+t^2)^2}\\
\cos{3x} &= 4\cos^3{x} - 3 \cos{x} = \frac{(1-t^2)^3}{(1+t^2)^3} - \frac{3(1-t^2)}{1+t^2}
\end{align}
Substituting in your equation, and up to possible errors, one finally gets the equation
$$
t^6 + 2t^5 -3t^4 -8t^3 +11t^2 +6t -1 = 0
$$
which has two real roots and four complex roots. According to WolframAlpha, the real roots are
$0.136187$ and $- 0.548975$.
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