Cos^2 (2x) + cos^2 (3x) = 1

Cos^2 (2x) + cos^2 (3x) = 1;

(2cos^2x - 1) ^2 + (4cos^3x - 3cosx) ^2 = 1;

4cos^4x - 4cos^2x + 1 + 16cos^6x - 24cos^4x + 9cos^2x = 1;

16cos^6x - 20cos^4x + 5cos^2x = 0;

cos^2x (16cos^4x - 20cos^2x + 5) = 0;

cos^2x = 0 или 16cos^4x - 20cos^2x + 5 = 0;

1) cosx = 0;

x = pi/2 + pin, n ∈ Z;

2) 16cos^4x - 20cos^2x + 5 = 0;

Пусть cos^2x = t, t ∈ [0; 1];

16t^2 - 20t + 5 = 0;

D = 400 - 4 * 16 * 5 = 400 - 320 = 80;

t = (20 + - sqrt80) / 32 = (5 + - sqrt5) / 8;

cos^2x = (5 + - sqrt5) / 8;

cosx = + - sqrt ((5 + - sqrt5) / 8);

x = + -arccos ( + - sqrt ((5 + - sqrt5) / 8)) + 2pin.

Ответ: pi/2 + pin, n ∈ Z или + -arccos ( + - sqrt ((5 + - sqrt5) / 8)) + 2pin, n ∈ Z.