Question

Rezolvati, va rog ...................

Answer 1

Parerea mea:

cos₂x - sinxcox + 4sin₂x = 2;

Dar cos₂x - sin₂x = 1 => 1 + 3sin₂x - sinxcosx = 2 => 3sin₂x - sinxcosx = 1;

Dar cos₂x - sin₂x = 1 => 3sin₂x - sinxcosx = sin₂x + cos₂x <=> 2sin₂x - sinxcosx - cos₂x = 0 <=> -cos(2x) + sin₂x -sinxcosx = 0;

Relatia e echivalenta cu -cos(2x) - 1/2sin(2x) + sin₂x = 0 <=> -2cos(2x) - sin(2x) + 2sin₂x = 0 <=>2sin₂x = 2cos(2x) - 2*(1 - sin₂x);

-2cos(2x) + 2 * [cos(2x) - 1 + sin₂x) + 2sinxcosx = 0;

-2 + 2sin₂x + 2sinxcosx = 0;

-2sin₂x - 2cos₂x + 2 sin₂x + 2sinxcosx = 0;

cosx*[-2 - 2sinx] = 0;

Cazul I:Cosx = 0 pentru x = {3/2π,1/2π}.

Cazul II; -2 - 2sinx = 0 => x = {3/2π}.

Prin notatia cu indice 2(Exemplu: sin₂x) am facut referinta la sin patrat de x.