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Given that 2sin^2 θ + sinθ - 1 = 0, find the two values for sin θ

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2sin² θ + sin θ - 1 = 0 (2sin θ - 1)(sin θ + 1) = 0
imagine \(\sin \theta = x\) so this thingy will become 2x^2 + x - 1 = 0 do you know how to solve for x there?
so much can you understand ?

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No I do not know how to solve for x.. What's next?
Now we can split that into two equations 2sin θ - 1 = 0 and sin θ + 1 = 0 this much understood ?
sinθ = 1/2 and sinθ = -1 ?
Solving the first one: 2sin θ - 1 = 0 sin θ = 1/2 θ = 30°, 150°
Second: sin θ + 1 = 0 sin θ = -1 θ = 270°
Okay thank you
The question is asking for only two values for theta though, which one would we choose?

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