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:

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

Второе уравнение не имеет решений.

cos(x) = 1

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