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Trả lời:
$a,$
$2\cos\bigg{(}\dfrac{\pi}{4}-x\bigg{)}-\sin2x=1$
$⇔2\cos\bigg{(}\dfrac{\pi}{4}-x\bigg{)}-1=\sin2x$
$⇔\cos\bigg{(}\dfrac{\pi}{2}-2x\bigg{)}=\sin2x$
$⇔\sin2x=\sin2x$ (Luôn đúng).
$b,$
$2\sin^2\bigg{(}x+\dfrac{\pi}{4}\bigg{)}-1=\sin2x$
$⇔1-2\sin^2\bigg{(}x+\dfrac{\pi}{4}\bigg{)}+\sin2x=0$
$⇔\cos\bigg{(}2x+\dfrac{\pi}{2}\bigg{)}+\sin2x=0$
$⇔\cos\bigg{[}\dfrac{\pi}{2}-(-2x)\bigg{]}+\sin2x=0$
$⇔\sin(-2x)+\sin2x=0$
$⇔\sin2x-\sin 2x=0$ (Luôn đúng).
