2) sin 2x - cos 2x = 1 2sin x*cos x - 2cos^2 x + 1 = 1 2cos x*(sin x - cos x) = 0 cos x = 0; x1 = pi/2 + pi*k sin x - cos x = 0; sin x = cos x; tg x = 1; x2 = pi/4 + pi*n В промежуток [-pi; pi/3] = [-12pi/12; 4pi/12] попадают корни x1 = pi/2 - pi = -pi/2; x2 = pi/4 - pi = -3pi/4; x3 = pi/4
3) sin(pi+x/2) + cos(pi+x) = 1 -sin(x/2) - cos x = 1 -sin(x/2) - (1 - 2sin^2(x/2)) = 1 Замена sin(x/2) = t 2t^2 - t - 2 = 0 D = 1 - 4*2(-2) = 1 + 16 = 17
с^18 *c^15 *c^2=c^35