Solving trig equation on the interval 0

Question

Solving trig equation on the interval 0<theta<2pi

anonymous · Answer

$$3-3\sin \theta =6\cos^2\theta$$

anonymous · Answer

I solved it to equal $$0=6\sin^2\theta+3\sin \theta-2$$ just need help factoring

astrophysics · Answer

Let sin theta = x, and solve as you would regularly than plug it back in after :)

anonymous · Answer

That's what I'm having problems with, the factoring part. Whether it's sin theta or x doesn't make a difference to me. I have a lot of problems factoring :(

mertsj · Answer

$$3-3\sin \theta =6(1-\sin ^2\theta)$$
$$3-3\sin \theta =6-6\sin ^2\theta $$
$$6\sin ^2\theta -3\sin \theta -3=0$$$$2\sin ^2\theta-\sin \theta-1=0$$

mertsj · Answer

$$(2\sin \theta+1)(\sin \theta-1)=0$$

mertsj · Answer

Set each factor equal to 0 and solve.

anonymous · Answer

ah I did $$3-3\sin \theta = 1+6\sin^2\theta$$ then got $$0=6\sin^2\theta+3\sin \theta -2$$

I didnt realize you were supposed to write it out like that which made the rest of it get messed up. Thanks.