Question

Giải phương trình: sin(3x - \(\dfrac{\pi}{4}\)) = sin2xsin(x + \(\dfrac{\pi}{4}\))

Answer 1

`sin(3x-\pi/4)=sin 2x sin(x+\pi/4)`

`<=>\sqrt{2}/2sin 3x-\sqrt{2}/2 cos 3x=sin 2x(\sqrt{2}/2sin x+\sqrt{2}/2 cos x)`

`<=>sin 3x-cos 3x=sin 2x sin x +sin 2x cos x`

`<=>sin 3x - cos 3x=1/2(cos x-cos 3x)+1/2(sin x +sin 3x)`

`<=>1/2sin 3x-1/2cos 3x=1/2cos x+1/2sin x`

`<=>1/\sqrt{2}sin 3x-1/\sqrt{2}cos 3x=1/\sqrt{2}cos x+1/\sqrt{2}sin x`

`<=>sin(3x-\pi/4)=sin(x+\pi/4)`

`<=>[(3x-\pi/4=x+\pi/4+k2\pi),(3x-\pi/4=[3\pi]/4-x+k2\pi):}`

`<=>[(x=\pi/4+k\pi),(x=\pi/4+k\pi/2):}`

`<=>x=\pi/4+k\pi/2`   `(k in ZZ)`