Решите уравнение : 2sin 2x + ^3=0 .
cos(-x/3)=^2/2.
tg(4x-1)=^3/3.
sin(4-3x/2)=-1.
ctg(-7x)=-1

Спасибо большое )

cos(-x/3)=^2/2.-x/3=+-π/4+2πnx=+-3π/4+6πntg(4x-1)=^3/3.4x-1=π/6+πn4x=1+π/6+πnx=1/4+π/24+πn/4sin(4-3x/2)=-1.4-3x/2=-π/2+2πn-3x/2=-4-π/2+2πnx=8/3+π/3+2πn/3ctg(-7x)=-1-7x=3π/4+πnx=-3π/28+πn/7