Question

решите уравнение. 
4sin(в 4 степени)2x+3cos4x-1=0

Answer 1

$4sin^4 2x + 3cos 4x - 1 = 0\\ left[ cos 2x = cos^2 x - sin^2 x right]\\ 4sin^4 2x + 3(cos^2 2x - sin^2 2x) - 1 = 0\\ 4sin^4 2x + 3cos^2 2x - 4sin^2 2x + sin^2 2x - 1 = 0\\ left[ sin^2x - 1 = -cos^2x right]\\ 4sin^2 2x(sin^2 2x - 1) + 3cos^2 2x - cos^2 2x = 0\\ - 4sin^2 2x cos^2 2x + 2 cos^2 2x = 0\\ 2cos^2x (1 - 2sin^2 2x) = 0$

$left[ 1 - 2sin^2 x = cos 2x right]\\ 2cos^2 2x cos 4x = 0\\ 1. cos 2x = 0\\ 2x = pi n + frac{pi}{2}, n in mathbb{Z}\\ boxed{x = frac{pi}{2}n + frac{pi}{4}, n in mathbb{Z}}\\ 2. cos 4x = 0\\ boxed{x = frac{pi}{4}n + frac{pi}{8}, n in mathbb{Z}}$