Trigonometric equation: sin(x)sin(4x) = 1

[Math Tutor]

Find the roots of the trigonometric equation:
sin(x)sin(4x)=1

[Jessica]

=> 2 cos(4x + x)/2 sin(4x - x)/2 = 0
=> cos 5x/2 = 0 or sin 3x/2 = 0
=> 5x/2 = 2kπ ± π/2 or 3x/2 = kπ
=> x = 4kπ/5 ± π/5 or x = 2kπ/3
=> x = { 0, π/5, 3π/5, 2π/3, π, 4π/3, 7π/5, 9π/5, 2π }(answers)




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[Guest]

Why sin(x)sin(4x)=1 <=>
2 cos(4x + x)/2 sin(4x - x)/2 = 0

It seems wrong.

[Guest]

sin x sin 4x = 1/2(cos 5x-cos 3x) = 1. Then cos 5x = 2+cos 3x. That's possible only if cos 5x = 1 and cos 3x = -1 but such x does not exist. Thus the equation has no solution.

[MM]

Actually it is [tex]sin x \sin 4x = \frac{(\cos 3x-\cos 5x)}{2} = 1[/tex]. Thus [tex]\cos 3x=1[/tex] leading to [tex]x=\frac{k\pi}{3}[/tex] where [tex]k\in \mathbb{Z}[/tex]. But obviously [tex]\cos \frac{5k\pi}{3}\ne -1[/tex]. Finally no solution!