# prove $\sin{x}+\sin{2x}+\sin{3x}=3$ does not hold for any real $x$? [closed]

How do we prove that the equality $$\sin{x}+\sin{2x}+\sin{3x}=3$$ does not hold for any real value of $x$ only using trigonometry ?

## closed as off-topic by Namaste, Claude Leibovici, user26857, Davide Giraudo, ArnaldoMay 7 '17 at 15:07

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$\sin(x)=1$ implies that $x=\pi/2+2\pi k$, this implies that $\sin(2x)=0$, so it is not true. Since you must have $\sin(x)=\sin(2x)=\sin(3x)=1$ since $\sin(x)\in [-1,1]$.
• how can we prove that the equality implies $\sin x =\sin 2x = \sin 3x = 1$ ? – ss1729 May 6 '17 at 19:36
• because $sin(x)\leq 1$ for every $x$ so $sin(2x)\leq 1, sin(3x)\leq 1$. So if one of the numbers $sin(x),sin(2x), sin(3x)$ is strictly inferior to $1$, their sum will be strictly inferior to $3$. – Tsemo Aristide May 6 '17 at 19:37
That would only be possible if all of $\sin x$, $\sin 2x$ and $\sin 3x$ equalled $1$. But $\sin x=1$ implies $\sin 2x=0$.
A good follow-up: find the maximum of $\sin x+\sin 2x+\sin3x$ for real $x$.