Question

Решите уравнения: 1)2cos(2x-п/3)-1=0 2)7sin^2x-9sinx*cosx=-1 3)sin2x+cos(6x-п/3)=0

Answer 1

1)2cos(2x-п/3)-1=0
cos(2x-π/3)=1/2
2x-π/3=-π/3+2πk U 2x-π/3=π/3+2πk
2x=2πk U 2x=2π/3+2πk
x=πk U x=π/3+2πk,k∈z

2)7sin^2x-9sinx*cosx=-1
7sin²x-9sinxcosx+sin²x+cos²x=0
8sin²x-9sinxcosx+cos²x=0/cos²x
8tg²x-9tgx+1=0
tgx=t
8t²-9t+1=0
D=81-32=49
t1=(9-7)/16=1/8⇒tgx=1/8⇒x=arctg1/8+⇒k,k∈z
t2=(9+7)/16=1⇒tgx=1⇒x=π/4+πk,k∈z

3)sin2x+cos(6x-п/3)=0
cos(π/2-2x)+cos(6x-π/3)=0
2cos(2x+π/12)*cos(4x-5π/12)=0
cos(2x+π/12)=0⇒2x+π/12=π/2+πk⇒2x=5π/12+πk⇒x=5π/24+πk/2,k∈z
cos(4x+5π/12)=0⇒4x+5π/12=π/2+πk⇒4x=π/24+πk⇒x=π/96+πk/4.k∈z