Can somebody explain to me how to solve this please?
tan2x=(cos2x)/2
$\displaystyle \tan(2x)=\frac{\sin(2x)}{\cos(2x)}=\frac{\cos(2x)} {2}$
cross multipliing we get
$\displaystyle 2\sin(2x)=\cos^2(2x)=1-\sin^2(2x)\Rightarrow{\sin^2(2x)+2\sin2x)-1=0}$
Now let $\displaystyle \sin(2x)=u$
giving us
$\displaystyle u^2+2u-1$
solving by quadratic equation we get
$\displaystyle \frac{-2\pm\sqrt{4-4(1)(1)}}{2}$
can you go from there?
Note, that once you get the answer for the quadratic, you must do the following
Let the solution of the qudratic be $\displaystyle u=\pm{c}$
since you want the answer for x, you go back to yoru sub
$\displaystyle u=\sin(2x)=\pm{c}\Rightarrow{x=\frac{\arcsin\pm{c} }{2}}$