Can somebody explain to me how to solve this please?

tan2x=(cos2x)/2

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- Jun 2nd 2008, 06:14 PMunmcintoshsolving trig equations
Can somebody explain to me how to solve this please?

tan2x=(cos2x)/2 - Jun 2nd 2008, 06:21 PMMathstud28
$\displaystyle \tan(2x)=\frac{\sin(2x)}{\cos(2x)}=\frac{\cos(2x)} {2}$

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? - Jun 2nd 2008, 06:24 PMunmcintosh
yes i can,

thank you! - Jun 2nd 2008, 06:30 PMMathstud28
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

$\displaystyle u=\sin(2x)=\pm{c}\Rightarrow{x=\frac{\arcsin\pm{c} }{2}}$ - Jun 2nd 2008, 08:48 PMChris L T521