solving trig equations

**unmcintosh** · June 2nd 2008, 06:14 PM

Can somebody explain to me how to solve this please?
tan2x=(cos2x)/2

**Mathstud28** · June 2nd 2008, 06:21 PM

Originally Posted by unmcintosh

Can somebody explain to me how to solve this please?
tan2x=(cos2x)/2

$\tan(2x)=\frac{\sin(2x)}{\cos(2x)}=\frac{\cos(2x)} {2}$

cross multipliing we get

$2\sin(2x)=\cos^2(2x)=1-\sin^2(2x)\Rightarrow{\sin^2(2x)+2\sin2x)-1=0}$

Now let $\sin(2x)=u$

giving us

$u^2+2u-1$

solving by quadratic equation we get

$\frac{-2\pm\sqrt{4-4(1)(1)}}{2}$

can you go from there?

**unmcintosh** · June 2nd 2008, 06:24 PM

yes i can,
thank you!

**Mathstud28** · June 2nd 2008, 06:30 PM

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 $u=\pm{c}$

since you want the answer for x, you go back to yoru sub

$u=\sin(2x)=\pm{c}\Rightarrow{x=\frac{\arcsin\pm{c} }{2}}$

**Chris L T521** · June 2nd 2008, 08:48 PM

Originally Posted by Mathstud28

$\tan(2x)=\frac{\sin(2x)}{\cos(2x)}=\frac{\cos(2x)} {2}$

cross multipliing we get

$2\sin(2x)=\cos^2(2x)=1-\sin^2(2x)\Rightarrow{\sin^2(2x)+2\sin2x)-1=0}$

Now let $\sin(2x)=u$

giving us

$u^2+2u-1$

solving by quadratic equation we get

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

can you go from there?

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

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

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