Solve $\sin(17x)+\sin(13x)=\sin(7x)$

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Solve $\sin(17x)+\sin(13x)=\sin(7x)$

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#1 h Apr 29, 2025, 1:06 AM

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How to solve the equation:
$\sin(17x)+\sin(13x)=\sin(7x),\;0<x<24^{\circ}$ Approach: I'm trying to factor $\sin(18x)$ to get $x=10^{\circ}$ .

Any hint would be appreciated.

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#2 h Apr 30, 2025, 1:47 PM

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We can let $u=\sin x$ and apply $\sin[(2n+1)\theta]=(-1)^nT_{2n+1}(\sin\theta)$ in terms of Chebyshev polynomials.

So we are solving $T_{17}(u)+T_{13}(u)+T_7(u)=0$ .

Applying the recursion $T_0(u)=1$ , $T_1(u)=u$ , $T_n(u)=2uT_{n-1}(u)-T_{n-2}(u)$ gives:

$T_7(u)=64u^7-112u^5+56u^3-7u$
$T_{13}(u)=4096 u^{13}-13312u^{11}+16640u^9-9984u^7+2912u^5-364u^3+13u$
$T_{17}(u)=65536u^{17}-278528u^{15}+487424u^{13}-452608u^{11}+239360u^9-71808u^7+11424u^5-816u^3+17u$

So we are solving $65536u^{17}-278528u^{15}+491520u^{13}-465920u^{11}+256000u^9-81728u^7+14224u^5-1124u^3+23u=0$

Factoring this gives:

$u(8u^3-6u-1)(8u^3-6u+1)(1024u^{10}-2816u^8+2880u^6-1360u^4+296u^2-23)=0$

In radians:
$u=0\implies\sin x=0\implies x\in\left\{\left.\pi k\right|k\in\mathbb{Z}\right\}$ .
$8u^3-6u\pm 1=0\implies\sin 3x=\pm\frac{1}{2}\implies x=\left\{\frac{2\pi k}{3}\pm\left.\frac{\pi}{18},\frac{2\pi k}{3}\pm\frac{5\pi}{18}\right|k\in\mathbb{Z}\right\}$

In the last factor, if you put $y=u^2$ , you have $1024y^5-2816y^4+2880y^3-1360y^2+296y-23$ with Galois Group $S_5$ owing to exactly three of five roots being real. So these answers must be approximated:
The solutions from this factor are as follows.

$x=2\pi k\pm 0.426706\ldots$
$x=2\pi k\pm 1.111480\ldots$
$x=2\pi k\pm 1.443943\ldots$
$x=2\pi k\pm 1.697648\ldots$
$x=2\pi k\pm 2.030112\ldots$
$x=2\pi k\pm 2.714885\ldots$
$x=2\pi k+0.682274\ldots\pm 0.098079\ldots i$
$x=2\pi k-0.682274\ldots\pm 0.098079\ldots i$
$x=2\pi k+2.451398\ldots\pm 0.098079\ldots i$
$x=2\pi k-2.451398\ldots\pm 0.098079\ldots i$

If you convert these to degrees, the only solutions in the $(0^\circ,24^\circ)$ interval is $10^\circ$ .

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