Решение задания приложено
F'(x) = 1,5(sin2x)' - 5(sinx)' - x' = 3cos2x - 5cosx - 1
f''(x) = 3(cos2x)' - 5(cosx)' -1' = -6sin2x + 5sinx = 0
5sinx - 6(2sinxcosx) = 0
5sinx - 12sinxcosx = 0
sinx(5-12cosx)=0
5-12cosx=0
12cosx=5
cosx=5/12=0.416
x = acos(0.416) = 65.4
√24=√4*6=2√6
√1/√6=1/√6=√6/6
3√2/√3=3√2*√3/√3*√3=3√6/3=√6
5√3/√2=5√6/2
3√6-2√6+√6/6+√6-5√6/2=2√6+√6/6-5√6/2=-√6/3