# trig question!

• Oct 1st 2008, 09:16 AM
snowball
trig question!
solve the equation:
root3*sin2 $\theta$+2sin^2 $\theta$=1 for 0< $\theta$< $\pi$

thanks!(Wink)
• Oct 1st 2008, 09:24 AM
Chris L T521
Quote:

Originally Posted by snowball
solve the equation:
root3*sin2 $\theta$+2sin^2 $\theta$=1 for 0< $\theta$< $\pi$

thanks!(Wink)

I'll nudge you in the right direction:

$\sqrt{3}\sin(2\vartheta)+2\sin^2\vartheta=1$

Note that $\cos(2\vartheta)=1-2\sin^2\vartheta$.

So we end up with $\sqrt{3}\sin(2\vartheta)=1-2\sin^2\vartheta\implies \sqrt{3}\sin(2\vartheta)=\cos(2\vartheta)$

Thus we end up with $\sqrt{3}=\cot(2\vartheta)$

Can you take it from here?

--Chris
• Oct 1st 2008, 12:12 PM
snowball
thanks,that's very helpful, i can take it from there (Clapping)