F(x)=3* 1/3 *sin3x+c=sin3x+c
F(x)=3*(1:1/2)(sinx/2+c=6sinx/2 +c
3(2cos^2(x) - 1) - 5cosx = 1
6cos^2(x) - 5cosx - 4 = 0
Замена: cosx = t, t∈[-1;1]
6t^2 - 5t - 4 = 0
D = 25 + 4*6*4 = 121
t1 = (5 - 11)/12 = -6/12 = -1/2
t2 = (5+11)/12 = 16/12 > 1 - посторонний корень
cosx = -1/2
x = 2π/3 + 2πk, k∈Z
x = 4π/3 + 2πk, k∈Z
или можно записать по-другому:
x = +-2π/3 + 2πk, k∈Z
Tgb(1+1/cos2b)=tgb•(1+cos2b)/cos2b=
tgb•2cos^2(b)/cos2b=sinb/cosb•2cos^2(b)/
cos2b=2sinb•cosb/cos2b=sin2b/cos2b=
tg2b