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

Question

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

Answer 1

cos(-x/3)=^2/2.
-x/3=+-π/4+2πn
x=+-3π/4+6πn

tg(4x-1)=^3/3.
4x-1=π/6+πn
4x=1+π/6+πn
x=1/4+π/24+πn/4

sin(4-3x/2)=-1.
4-3x/2=-π/2+2πn
-3x/2=-4-π/2+2πn
x=8/3+π/3+2πn/3

ctg(-7x)=-1
-7x=3π/4+πn
x=-3π/28+πn/7

Comments

Спасибо большое )