Solve the equation $$\sin \theta = 2 \cos 2 \theta +1--------0\le \theta \le 360$$

Question

Solve the equation $$\sin \theta = 2 \cos 2 \theta +1-------->0\le \theta \le 360$$

anonymous · Answer

sinx=2cos2x +1
sinx=2[cos ^2 x - sin ^2 x]+1
sinx=2[1-sin^2 x-sin^2 x]+1
sinx=2+1-4sin^2 x
4sin^2 x+sinx-3=0
4sin^2 x +4sinx-3sinx-3=0
4sinx(sinx+1)-3(sinx+1)=0
(4sinx-3)(sinx+1)=0
sinx=3/4   or   sinx =-1
check it!!!

anonymous · Answer

oh sorry it was theta not x...

anonymous · Answer

very nice...

anonymous · Answer

I got to the bit where sinx=2+1-4sin^2 x, but i forgot how to carry on from there...

anonymous · Answer

http://www.wolframalpha.com/input/?i=sinx%3D2cos2x+%2B1

anonymous · Answer

just arrange it to the form of a quadratic equation in sinx

anonymous · Answer

did you get it @order???

anonymous · Answer

Yes, I'd just forgotten I could do that.