Question

Решите уравнение: a) sin2x+√3 cos2x=cos(п/6-2x). б)Найдите корни удовлетворяющие условию 0<x<4п/3.

Answer 1

Sin2x+√3cos2x=cosπ/6cos2x+sinπ/6sin2x
sin2x+√3cos2x-√3/2cos2x-1/2sin2x=0/82
2sin2x+2√3cos2x-√3cos2x-sin2x=0
sin2x+√3cos2x=0/cos2x
tg2x+√3=0
tg2x=-√3
2x=-π/3+πk
x=-π/6+πk/2,k∈z
0<-π/6+πk/2<4π/3  /*(6/π)
0<-1+3k<8
1<3k<9
1/3<k<3
k=1⇒x=-π/6+π/2=π/3
k=2⇒x=-π/6+π=5π/6