# Thread: trig question!

1. ## trig question!

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

thanks!

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

thanks!
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

3. thanks,that's very helpful, i can take it from there