solve 3 cos (2y )- 5 cos( y)+ 2 =0 where 0<y<360 degrees

cos(2y)= (cos(y))^2 - (sin(y))^2 identitynow consider identity (cos(y))^2+(sin(y))^2=1by rearranging this to (cos(y))^2 - 1 = - (sin(y))^2 we can now substitute it into our first identity to obtaincos(2y)= 2(cos(y))^2 - 1 We can now substitute cos(2y) in our equation to be solved.3(2(cos(y))^2 - 1) -5cos(y)+2=0simplifying this, we get:6(cos(y))^2 -3 -5cos(y) +2=06(cos(y))^2 -5cos(y) -1=0now let cos(y)=x, and solve this equation as a quadratic through factorisation6x² -5x-1=06x² -6x +x-1=06x(x-1)+ (x-1)=0(6x+1)(x-1)=0so x=-(1/6) or x=1so cos(y)=-(1/6) or cos(y)=1so y=pi/2or arccos(-1/6)=y