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Question Number 13842 by Tinkutara last updated on 24/May/17

Find the number of solutions of the equation sin 5x cos 3x = sin 6x cos 2x, x ∈ [0, π]

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{sin}\:\mathrm{5}{x}\:\mathrm{cos}\:\mathrm{3}{x}\:=\:\mathrm{sin}\:\mathrm{6}{x}\:\mathrm{cos}\:\mathrm{2}{x}, \\ $$$${x}\:\in\:\left[\mathrm{0},\:\pi\right] \\ $$

Answered by myintkhaing last updated on 24/May/17

five solutions

$${five}\:{solutions} \\ $$

Answered by myintkhaing last updated on 24/May/17

2 sin 5x cos 3x = 2 sin 6x cos 2x sin 8x + sin 2x = sin 8x + sin 4x 2 sin 2x cos 2x − sin 2x = 0 sin 2x (2cos 2x − 1) = 0 sin 2x = 0 or cos 2x = (1/2) 0 ≤ 2x ≤ 2π 2x = 0, π, 2π or 2x = (π/3), ((5π)/3) x = 0 , (π/2), π or (π/6), ((5π)/6)

$$\mathrm{2}\:{sin}\:\mathrm{5}{x}\:{cos}\:\mathrm{3}{x}\:=\:\mathrm{2}\:{sin}\:\mathrm{6}{x}\:{cos}\:\mathrm{2}{x} \\ $$$${sin}\:\mathrm{8}{x}\:+\:{sin}\:\mathrm{2}{x}\:=\:{sin}\:\mathrm{8}{x}\:+\:{sin}\:\mathrm{4}{x} \\ $$$$\mathrm{2}\:{sin}\:\mathrm{2}{x}\:{cos}\:\mathrm{2}{x}\:−\:{sin}\:\mathrm{2}{x}\:=\:\mathrm{0} \\ $$$${sin}\:\mathrm{2}{x}\:\left(\mathrm{2}{cos}\:\mathrm{2}{x}\:−\:\mathrm{1}\right)\:=\:\mathrm{0} \\ $$$${sin}\:\mathrm{2}{x}\:=\:\mathrm{0}\:{or}\:{cos}\:\mathrm{2}{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{0}\:\leqslant\:\mathrm{2}{x}\:\leqslant\:\mathrm{2}\pi \\ $$$$\mathrm{2}{x}\:=\:\mathrm{0},\:\pi,\:\mathrm{2}\pi\:{or}\:\mathrm{2}{x}\:=\:\frac{\pi}{\mathrm{3}},\:\frac{\mathrm{5}\pi}{\mathrm{3}} \\ $$$${x}\:=\:\mathrm{0}\:,\:\frac{\pi}{\mathrm{2}},\:\pi\:{or}\:\frac{\pi}{\mathrm{6}},\:\frac{\mathrm{5}\pi}{\mathrm{6}} \\ $$

Commented by Tinkutara last updated on 24/May/17

Thanks.

$$\mathrm{Thanks}. \\ $$

Answered by mrW1 last updated on 24/May/17

sin 5x cos 3x = sin 6x cos 2x sin 5x cos 3x = 2sin 3x cos 3x cos 2x (1) cos 3x=0 3x=(π/2),((3π)/2),((5π)/2) ⇒x=(π/6),(π/2),((5π)/6) (2) cos 3x≠0 sin 5x = 2sin 3x cos 2x sin 3x cos 2x+cos 3x sin 2x= 2sin 3x cos 2x sin 3x cos 2x−cos 3x sin 2x= 0 sin (3x−2x)= 0 sin x = 0 ⇒x=0,π

$$\mathrm{sin}\:\mathrm{5}{x}\:\mathrm{cos}\:\mathrm{3}{x}\:=\:\mathrm{sin}\:\mathrm{6}{x}\:\mathrm{cos}\:\mathrm{2}{x} \\ $$$$\mathrm{sin}\:\mathrm{5}{x}\:\mathrm{cos}\:\mathrm{3}{x}\:=\:\mathrm{2sin}\:\mathrm{3}{x}\:\mathrm{cos}\:\mathrm{3}{x}\:\mathrm{cos}\:\mathrm{2}{x} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{cos}\:\mathrm{3}{x}=\mathrm{0} \\ $$$$\mathrm{3}{x}=\frac{\pi}{\mathrm{2}},\frac{\mathrm{3}\pi}{\mathrm{2}},\frac{\mathrm{5}\pi}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\frac{\pi}{\mathrm{6}},\frac{\pi}{\mathrm{2}},\frac{\mathrm{5}\pi}{\mathrm{6}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{cos}\:\mathrm{3}{x}\neq\mathrm{0} \\ $$$$\mathrm{sin}\:\mathrm{5}{x}\:=\:\mathrm{2sin}\:\mathrm{3}{x}\:\mathrm{cos}\:\mathrm{2}{x} \\ $$$$\mathrm{sin}\:\mathrm{3}{x}\:\mathrm{cos}\:\mathrm{2}{x}+\mathrm{cos}\:\mathrm{3}{x}\:\mathrm{sin}\:\mathrm{2}{x}=\:\mathrm{2sin}\:\mathrm{3}{x}\:\mathrm{cos}\:\mathrm{2}{x} \\ $$$$\mathrm{sin}\:\mathrm{3}{x}\:\mathrm{cos}\:\mathrm{2}{x}−\mathrm{cos}\:\mathrm{3}{x}\:\mathrm{sin}\:\mathrm{2}{x}=\:\mathrm{0} \\ $$$$\mathrm{sin}\:\left(\mathrm{3}{x}−\mathrm{2}{x}\right)=\:\mathrm{0} \\ $$$$\mathrm{sin}\:{x}\:=\:\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{0},\pi \\ $$

Commented by Tinkutara last updated on 24/May/17

Thanks.

$$\mathrm{Thanks}. \\ $$

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