Question

решить уравнение,помогите ​

Answer 1

Ответ: π/4 + πk, k є Z

Объяснение:

cos^2 (2x) = 1 - sin^2 (2x)

1 + sin2x - 2cos^2 (2x) = 2sin^2 (2x)

1 + sin2x - 2 + 2sin^2 (2x) = 2sin^2 (2x)

sin 2x = 1

2x = π/2 + 2πk, k є Z

x = π/4 + πk, k є Z

Answer 2

$1+Sin2x-2Cos^{2}2x=2Sin^{2}2x\\\\1+Sin2x-2Cos^{2}2x-2Sin^{2}2x=0\\\\1+Sin2x-2(Cos^{2}2x+Sin^{2}2x)=0\\\\1+Sin2x-2=0\\\\Sin2x-1=0\\\\Sin2x=1\\\\2x=\frac{\pi }{2}+2\pi n,n\in Z\\\\x=\frac{\pi }{4}+\pi n,n\in Z$