Question

(√2)*(sin2x-cos2x)=cos4x

P.S: sin2x это именно sin2x, а не sin²x

Answer 1

√2(Sin2x - Cos2x) = Cos4x

√2(Sin2x - Cos2x) = Cos²2x - Sin²2x

-√2( Cos2x - Sin2x) -( Cos2x - Sin2x)(Cos2x + Sin2x) = 0

(Cos2x - Sin2x)(√2 - Cos2x - Sin2x) = 0

1) Cos2x - Sin2x = 0 |: Cos2x , Cos2x ≠ 0

1 - tg2x = 0

tg2x = 1

$2x=arctg1+\pi n,n\in Z\\\\2x=\frac{\pi }{4}+\pi n,n\in Z\\\\\boxed{x=\frac{\pi }{8}+\frac{\pi n}{2},n\in Z}$

$2)\sqrt{2}-Cos2x-Sin2x=0\\\\Cos2x+Sin2x=\sqrt{2} |:2\\\\\frac{1}{2}Cos2x+\frac{1}{2} Sin2x=\frac{\sqrt{2}}{2}\\\\Cos\frac{\pi }{6}Cos2x+Sin\frac{\pi }{6}Sin2x=\frac{\sqrt{2} }{2} \\\\Cos(2x-\frac{\pi }{6}=\frac{\sqrt{2} }{2}\\\\2x-\frac{\pi }{6}=\pm arc Cos\frac{\sqrt{2} }{2}+2\pi n,n\in Z\\\\2x-\frac{\pi }{6}=\pm \frac{\pi }{4} +2\pi n,n\in Z\\\\2x=\pm \frac{\pi }{4}+\frac{\pi }{6} +2\pi n,n\in Z\\\\\boxed{x=\pm \frac{\pi }{8}+\frac{\pi }{12} +\pi n,n\in Z}$