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Question Number 151248 by mathdanisur last updated on 19/Aug/21

	Answered by EDWIN88 last updated on 19/Aug/21

	$4 \sin (4 x - 60 °) \sin (6 x - 60 °) \sin (480 ° - 10 x) + \frac{\sqrt{3}}{2} = 0$ ${2 \sin (4 x - 60 °) \sin (6 x - 60 °)} 2 \sin (120 ° - 10 x) + \sin 60 ° = 0$ ${\cos 2 x - \cos (10 x - 120 °)} 2 \sin (120 ° - 10 x) + \sin 60 ° = 0$ $2 \cos 2 x \sin (120 ° - 10 x) - \sin (240 ° - 20 x) + \sin 60 ° = 0$ $\sin (120 ° - 8 x) - \sin (12 x - 120 °) + \sin (60 ° - 20 x) + \sin 60 ° = 0$ $\sin (120 ° - 8 x) + \sin 60 ° + \sin (60 ° - 20 x) - \sin (12 x - 120 °) = 0$ $2 \sin (90 ° - 4 x) \cos (30 ° - 4 x) + 2 \cos (30 ° - 4 x) \sin (90 ° - 16 x) = 0$ $2 \cos 4 x \cos (30 ° - 4 x) + 2 \cos (30 ° - 4 x) \cos 16 x = 0$ $2 \cos (30 ° - 4 x) {\cos 16 x + \cos 4 x} = 0$ $4 \cos (30 ° - 4 x) \cos 10 x \cos 6 x = 0$ ${\begin{cases} \cos (30 ° - 4 x) = 0 \\ \cos 10 x = 0 \\ \cos 6 x = 0 \end{cases}$ ${\begin{cases} \cos (4 x - 30 °) = \cos 90 ° \Rightarrow 4 x = 30 ° \pm 90 ° + 2 k π \\ \cos 10 x = \cos 90 ° \Rightarrow x = \pm 9 ° + k . 36 ° \\ \cos 6 x = \cos 90 ° \Rightarrow x = \pm 15 ° + k . 60 ° \end{cases}$
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