Question

4 cos x - sin ^2 -4=0

Answer 1

$4cos(x)-sin^2(x) - 4 = 0\4cos(x) - sin^2(x) - 4 + 1 - 1 = 0\4cos(x) - sin^2(x) - 5 + 1 = 0 | 1 = cos^2(x) + sin^2(x) |\4cos(x)+cos^2(x) - 5 = 0$

пусть cos²(x) = t, тогда:

$4t +t^2-5 = 0\t^2 + 4t-5=0\D = 4*4 -4*(-5)=36\t = frac{-4pm6}{2} \cos(x_1) = 1\cos(x_2) = 5$

Второе уравнение не имеет решений.

cos(x) = 1

$x=pm arccos(1) + 2pi * k \x = 0+2pi * k\x = 2pi*k kin Z$