I know I need to set \(\displaystyle cos(2\alpha)=1-2\sin^2(\alpha)\) which is then \(\displaystyle \frac{31}{49}=1-2\sin^2(\alpha)\) which is then \(\displaystyle 2\sin^2(\alpha)=1-\frac{31}{49}\) now I think I get \(\displaystyle 2\sin^2(\alpha)= \frac{49}{49}-\frac{39}{49}=\frac{18}{49}\) from here I need some help ?\(\displaystyle 2\sin^2(\alpha)=\frac{18}{49}\)