# Unit Circle-Again

• Nov 2nd 2008, 05:39 PM
Stevo_Evo_22
Unit Circle-Again
Hey everyone,

I really appreciated your help last time, now I have revision for an exam:(

Find values for x between -pi and pi if sin2x=root3/2

I know that sin(pi/3) = root3/2. I have been able to obtain values of pi/3 and pi/6, but the answers say that there are 2 more. I can't see the logic behind it, except I do know that you have to increase the initial range or something because it is 2x. I think.

Steven
• Nov 3rd 2008, 11:33 AM
Moo
Hello,
Quote:

Originally Posted by Stevo_Evo_22
Hey everyone,

I really appreciated your help last time, now I have revision for an exam:(

Find values for x between -pi and pi if sin2x=root3/2

I know that sin(pi/3) = root3/2. I have been able to obtain values of pi/3 and pi/6, but the answers say that there are 2 more. I can't see the logic behind it, except I do know that you have to increase the initial range or something because it is 2x. I think.

Steven

Draw a circle.

$\tfrac{\sqrt{3}}{2}=\left\{\begin{array}{ll} \sin \tfrac \pi 3 \\ \sin \pi-\tfrac \pi 3=\sin \tfrac{2 \pi}{3} \end{array} \right.$

Now, if $\sin(a)=\sin(b)$, look at your circle :
$a=\pi-b+2k \pi$

Now, you're correct to point out the initial range.
If $x \in [-\pi,\pi ]$, then $y=2x \in [-2 \pi, 2 \pi]$

so you'll have to solve for y in $\sin y=\sqrt{3}/2$ in that new domain !
• Nov 3rd 2008, 04:23 PM
Stevo_Evo_22
Thanks! So would 'a' (pi/3) be pi + pi/3? Because we've already gone round a half time?