Question

2(cosx-1)sin2x=3cos(3п/2+x)
[3п/2;3п]​

Answer 1

$2(cos{x}-1)2sin{x}cos{x}-3cos{(3pi/2+x)}=0\cos{(3pi/2+x)}=sin{x}\4sin{x}(cos^2{x}-cos{x}-3/4)=0\D=1+3=2^2\sin{x}(cos{x}+1/2)(cos{x}-3/2)=0\!|cos{x}|leq 1!\left[begin{array}{ccc}sin{x}=0\cos{x}=-1/2\end{array}$

x={pi*n;±2pi/3+2pi*n}, n∈Z.

Ответ: x={2pi; 8pi/3; 3pi}