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anonymous

  • one year ago

what steps would I take to solve the problem: Solve each equation on the interval 0<alpha<2pie 1)2sinx+sqrt2=0 2)2sin^2x-3sinx+1=0

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  1. anonymous
    • one year ago
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    Do you mean \(0<x<2\pi\) or do you mean \(0 < \alpha < 2\pi\)?

  2. anonymous
    • one year ago
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    the 2nd one

  3. anonymous
    • one year ago
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    But there are no alphas in your problems...

  4. anonymous
    • one year ago
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    that why im confused as well lol maybe it is the first one

  5. anonymous
    • one year ago
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    Let's just assume that it is the first one, because that would make more sense lol

  6. anonymous
    • one year ago
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    So, for the first one.\[2\sin(x)+\sqrt{2}=0\]\[2\sin(x)=-\sqrt{2}\]\[\sin(x)=-\frac{\sqrt{2}}{2}\]Now, use your trig tables to find what values of x satisfy that while making sure \(0 < x < 2\pi\). Get it?

  7. anonymous
    • one year ago
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    oh so you pretty much want to get the sin alone

  8. anonymous
    • one year ago
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    what if there is sin and cos in the problem. ex:8-12sin^2x=4cos^2x

  9. anonymous
    • one year ago
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    Yeah. When solving trig equations, you want to (1) get all the trig stuff into 1 trigonometric function. So, if you have both a sine and a cosine or something, you want to (usually) get it all in terms of sine or cosine. (2) You want to solve for the trig function.

  10. anonymous
    • one year ago
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    For that kind of thing, that's where you have to use your identities. For instance, in that question you just posted, trying using your pythagorean identities. \[\implies8 - 12\sin^2(x)=4\cos^2(x)\]\[\implies8-12\sin^2(x)=4(1-\sin^2(x))\]\[\implies0=8\sin^2(x)-4\]\[\implies\frac12=\sin^2(x)\]Does this make sense?

  11. anonymous
    • one year ago
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    ahh yes I finally understand :D thanks so much!

  12. anonymous
    • one year ago
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    You're welcome. :)

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