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Category: Algebra

Question-151248

Question Number 151248 by mathdanisur last updated on 19/Aug/21 Answered by EDWIN88 last updated on 19/Aug/21 4sin(4x60°)sin(6x60°)sin(480°10x)+32=0{2sin(4x60°)sin(6x60°)}2sin(120°10x)+sin60°=0{cos2xcos(10x120°)}2sin(120°10x)+sin60°=02cos2xsin(120°10x)sin(240°20x)+sin60°=0$$\mathrm{sin}\:\left(\mathrm{120}°−\mathrm{8}{x}\right)−\mathrm{sin}\:\left(\mathrm{12}{x}−\mathrm{120}°\right)+\mathrm{sin}\:\left(\mathrm{60}°−\mathrm{20}{x}\right)+\mathrm{sin}\:\mathrm{60}°=\mathrm{0}…