Question

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

Answer 1

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!!!

Answer 2

oh sorry it was theta not x...

Answer 3

very nice...

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

did you get it @order???

Answer 8

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