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

Question

Accepted Answer

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.

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