## anonymous one year ago Solving trig equation on the interval 0<theta<2pi

1. anonymous

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

2. anonymous

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

3. Astrophysics

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

4. anonymous

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 :(

5. Mertsj

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

6. Mertsj

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

7. Mertsj

Set each factor equal to 0 and solve.

8. anonymous

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.

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