Решите
tg(arcos(-√2/2);
1 + cosx = 2sin^2x;

2sin^2x+cosx-1=0(sin^2x +cos^2x = 1 => sin^2x = 1 - cos^2x)2*(1 - cos^2x) +cosx-1=02cos^2x-cosx-1 = 0t = cosx, t∈[-1;1]2t^2 - t - 1 = 0D = 1+8=9t1= (1+3)/4=1t2= (1-3)/4=-0.5cosx=1 cosx=-0.5xn= 2Пn, n∈Z xk=±arccos(-0.5) + 2Пk, k∈Z xk=±2П/3 + 2Пk, k∈ZОтвет: xn= 2Пn, n∈Z; xk=±2П/3 + 2Пk, k∈Z