Ответ:

Объяснение:2sinx/2·cosx/2+3(cos²x/2-sin²x/2)=2(sin²x/2+cos²x/2);

cos²x/2-5sin²x/2+2sinx/2·cosx/2=0;

5sin²x/2-2sinx/2·cosx/2-cos²x/2=0 ║ : cos²x/2≠0,

5tg²x/2-2tg x/2-1=0;  tg x/2=t,

5t²-2t-1=0, t1=(1+√6)/5 , t2=(1-√6)/5;

1)tg x/2=(1+√6)/5,  x/2=arc tg(1+√6)/5+πn,n∈z    x=2arc tg(1+√6)/5 +2πn,n∈z

2)tg x/2=(1-√6)/5,   x/2=arc tg(1-√6)/5 +πk,k∈z,    x=2arc tg(1-√6)/5 +2πk,k∈z