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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?

Blue
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ABC
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  • You need to have an equation where one side is zero. Your last line does not achieve anything because $0.1 \times 10 = 0.01 \times 100 = 0.001 \times 1000 = 1$: the two brackets could be anything. – Toby Mak Nov 21 '19 at 12:10
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    Where is this question from? I don't see any particular reason why one would expect this to have a "nice" solution, let alone solvable analytically – Brenton Nov 21 '19 at 12:16
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    I can get the roots on a period numerically, but division by $\pi$ doesn't give them obvious closed forms. – J.G. Nov 21 '19 at 12:26

2 Answers2

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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 $- 0.548975$.

J.-E. Pin
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Using https://www.wolframalpha.com/input/?i=1%2Bsin%28x%29%2Bsin%282x%29%2Bsin%283x%29%3Dcos%28x%29%2Bcos%282x%29%2Bcos%283x%29

It looks like there are two real periodic solutions and 3 more complex ones.

V-X
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