# solving trig equations

• Jun 2nd 2008, 06:14 PM
unmcintosh
solving trig equations
Can somebody explain to me how to solve this please?
tan2x=(cos2x)/2
• Jun 2nd 2008, 06:21 PM
Mathstud28
Quote:

Originally Posted by unmcintosh
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?
• Jun 2nd 2008, 06:24 PM
unmcintosh
yes i can,
thank you!
• Jun 2nd 2008, 06:30 PM
Mathstud28
Quote:

Originally Posted by unmcintosh
yes i can,
thank you!

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 PM
Chris L T521
Quote:

Originally Posted by Mathstud28
$\displaystyle \tan(2x)=\frac{\sin(2x)}{\cos(2x)}=\frac{\cos(2x)} {2}$

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

$\displaystyle \color{red}\boxed{\frac{-2\pm\sqrt{4-4(1)(1)}}{2}}$

can you go from there?

Inside the square root, it should be $\displaystyle \sqrt{4-4(1)(-1)}$.

The way you did it would give one solution, when in fact there are two solutions.