A calculus problem by Vu Hoang

Set $y=\cos{x}$ . Hence, the domain for y is $[0,1]$ . We have $4y^3-12y^2-19y+12=0$ . Factor out the equation $2(y-\frac{1}{2})(y+\frac{3}{2})(y-4)=0$ . Then $y=\frac{1}{2}$ or $x=\frac{π}{6}$ . Plug the value of x into 2nd equation $f(n)=\sin{\frac{π}{6}}+\sin{\frac{2π}{6}}+\sin{\frac{3π}{6}}+...+\sin{\frac{nπ}{6}}$ or $f(n)=\frac{\sqrt(3)}{2}+\frac{\sqrt(3)}{2}+1-\frac{\sqrt(3)}{2}-\frac{\sqrt(3)}{2}-1+...+\sin{\frac{nπ}{6}}=\sqrt(3)$ . It is just right for the first 2 terms of the function, so $n=2$