sin(9x) = sin(5x) + sin(3x) find: x = ?


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Question Number 211657 by hardmath last updated on 15/Sep/24

$\sin (9 x) = \sin (5 x) + \sin (3 x)$ $find : x = ?$
	Answered by Frix last updated on 15/Sep/24
	$sin ((2n+1)x) =sin x (1+2Σ_(k=1) ^n cos (2kx)) ⇒ sin 9x −sin 5x −sin 3x =0 ⇔ 2(cos 8x +cos 6x −cos 2x −(1/2))sin x =0 Let 0≤x<2π ★ x=0∨x=π cos 8x +cos 6x −cos 2x −(1/2)=0 This has a period of π and it′s symmetric to ((nπ)/2) ⇒ We must look for solutions x∈[0; (π/2)] These should be at (((2k+1)π)/(30)) Testing leads to ★ x=(π/(30)), ((7π)/(30)), ((11π)/(30)), ((13π)/(30)) [0; (π/2)] ⇒ Due to symmetry ★ x=((17π)/(30)), ((19π)/(39)), ((23π)/(30)), ((29π)/(30)) [(π/2); π] ★ All these +π [π; 2π) We get x=(π/(30)){0, 1, 7, 11, 13, 17, 19, 23, 29}+nπ$
	$\sin ((2 n + 1) x) = \sin x (1 + 2 \underset{k = 1}{\sum^{n}} \cos (2 k x))$ $\Rightarrow$ $\sin 9 x - \sin 5 x - \sin 3 x = 0$ $\Leftrightarrow$ $2 (\cos 8 x + \cos 6 x - \cos 2 x - \frac{1}{2}) \sin x = 0$ $Let 0 ⩽ x < 2 π$ $★ x = 0 \lor x = π$ $\cos 8 x + \cos 6 x - \cos 2 x - \frac{1}{2} = 0$ $This has a period of π and {it}^{'} s symmetric$ $to \frac{n π}{2}$ $\Rightarrow We must look for solutions x \in [0; \frac{π}{2}]$ $These should be at \frac{(2 k + 1) π}{30}$ $Testing leads to$ $★ x = \frac{π}{30}, \frac{7 π}{30}, \frac{11 π}{30}, \frac{13 π}{30} [0; \frac{π}{2}]$ $\Rightarrow Due to symmetry$ $★ x = \frac{17 π}{30}, \frac{19 π}{39}, \frac{23 π}{30}, \frac{29 π}{30} [\frac{π}{2}; π]$ $★ All these + π [π; 2 π)$ $We get$ $x = \frac{π}{30} {0, 1, 7, 11, 13, 17, 19, 23, 29} + n π$
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