2cosx(2sinx+1) =(2sinx+1)(2sinx-1) ;
2cosx(2sinx+1) -(2sinx+1)(2sinx-1) =0;
(2sinx+1)(2cosx -2sinx+1) =0 ;
a) 2sinx+1 =0 ;
sinx = -1/2 ;
x=(-1)^(n+1)*π/6 +π*n ,n∈Z.
б) 2cosx -2sinx+1 =0 ;
sinx -cosx =1/2 ;
√2sin(x-π/4) =1/2;
x-π/4 = (-1)^n*arcsin1/(2√2)+π*n, n∈Z.
x = π/4+ (-1)^n*arcsin1/(2√2)+π*n, n∈Z.