Question

Во вложении.................

Answer 1

a)cos(6t+π/2)=-√3/2
-sin6t=-√3/2
sin6t=√3/2
[6t=π/3+2πk⇒t=π/18+πk/3
[6t=2π/3+2πk⇒t=π/9+πk/3
б)-2sinxcos2x/(1+cosx)=0
1+cosx≠0⇒cosx≠-1⇒x≠π+2πk-ОДЗ
sinx=0⇒x=πk+ОДЗ⇒x=2πk
cos2x=0⇒2x=π/2+πk⇒x=π/4+πk/2
в)
4sin²3x+4cos²3x-cos²3x-3sin²3x-4sin3xcos3x=0
sin²3x-4sin3xcos3x+3cos²3x=0/cos²3x
tg²3x-4tg3x+3=0
tg3x=a
a²-4a+3=0
{a1+a2=4
{a1*a2=3
a1=1⇒tg3x=1⇒3x=π/4+πk⇒x=π/12+πk/3
a2=3⇒tg3x=3⇒3x=arctg3+πk⇒x=1/3*arctg3+πk/3