Question

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

Answer 1

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

x={pi*n;±2pi/3+2pi*n}, n∈Z.

Ответ: x={2pi; 8pi/3; 3pi}