## 4meisu 4 years ago Given that 2sin^2 θ + sinθ - 1 = 0, find the two values for sin θ

1. anonymous

2sin² θ + sin θ - 1 = 0 (2sin θ - 1)(sin θ + 1) = 0

2. anonymous

imagine $$\sin \theta = x$$ so this thingy will become 2x^2 + x - 1 = 0 do you know how to solve for x there?

3. anonymous

so much can you understand ?

4. anonymous

@4meisu ?

5. 4meisu

No I do not know how to solve for x.. What's next?

6. anonymous

Now we can split that into two equations 2sin θ - 1 = 0 and sin θ + 1 = 0 this much understood ?

7. 4meisu

yes

8. 4meisu

sinθ = 1/2 and sinθ = -1 ?

9. anonymous

Solving the first one: 2sin θ - 1 = 0 sin θ = 1/2 θ = 30°, 150°

10. anonymous

Second: sin θ + 1 = 0 sin θ = -1 θ = 270°

11. 4meisu

Okay thank you

12. anonymous

:)

13. 4meisu

The question is asking for only two values for theta though, which one would we choose?