Question

помогите 3-3cosx=2sin^2x

Answer 1

3 - 3cos(x) = 2sin^2(x);

3 - 3cos(x) = 2 - 2cos^2(x);

2cos^2(x) - 3cos(x) +1 = 0;

Пусть cos(x) = t, tE[-1;1], тогда:

2t^2 - 3t +1 = 0;

D = 1

x1 = 2 (посторонний корень);

x2 = 1.

cos(x) = 1;

x1,2 = +/- arccos 1 + 2piK, kEz

arccos 1 = 0;

Ответ: x = 2piK, KEz.