Find the solution of equation sin^2 x+sin^2 2x+sin^2 3x = 1 on interval (0,2π)


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Question Number 137366 by liberty last updated on 02/Apr/21

$F i n d t h e s o l u t i o n o f e q u a t i o n$ $\sin^{2} x + \sin^{2} 2 x + \sin^{2} 3 x = 1$ $o n i n t e r v a l (0, 2 π)$
	Answered by benjo_mathlover last updated on 02/Apr/21

	$(*) \sin^{2} x + \sin^{2} 3 x = 1 - \sin^{2} 2 x$ ${(\sin 3 x + \sin x)}^{2} - 2 \sin 3 x \sin x = \cos^{2} 2 x$ ${(2 \sin 2 x \cos x)}^{2} + \cos 4 x - \cos 2 x = \cos^{2} 2 x$ $4 \sin^{2} 2 x (\frac{1}{2} + \frac{1}{2} \cos 2 x) + \cos 4 x - \cos 2 x = \cos^{2} 2 x$ $l e t \cos 2 x = a$ $(1 - a^{2}) (2 + 2 a) + 2 a^{2} - 1 - a = a^{2}$ $2 + 2 a - 2 a^{2} - 2 a^{3} + 2 a^{2} - 1 - a - a^{2} = 0$ $- 2 a^{3} - a^{2} + a + 1 = 0$ $2 a^{3} + a^{2} - a - 1 = 0$ $x_{1} \approx 0.2963$ $x_{2} \approx 2.8453$ $x_{3} \approx 3.4379$ $x_{4} \approx 5.9869$
	Answered by liberty last updated on 02/Apr/21

	$\Leftrightarrow \frac{1 - \cos 2 x}{2} + \frac{1 - \cos 4 x}{2} = \cos^{2} 3 x$ $\frac{2 - \cos 2 x - \cos 4 x}{2} = \frac{1 + \cos 6 x}{2}$ $\cos 6 x + \cos 4 x + \cos 2 x = 1$ $\cos 4 x + 2 \cos 4 x \cos 2 x = 1$ ${2 \cos}^{2} 2 x - 1 + 2 \cos 2 x (2 \cos^{2} 2 x - 1) - 1 = 0$ $l e t \cos 2 x = c$ $\Rightarrow 2 c^{2} + 2 c (2 c^{2} - 1) - 2 = 0$ $\Rightarrow 2 c^{3} + c^{2} - c - 1 = 0$
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