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Babynini

  • one year ago

Solve the given equation 2cos^2(theta)+sin(theta)=1

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  1. Babynini
    • one year ago
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    I've gotten up to 2sin^2(theta)+sin(theta)+1=0 but cann't figure out how to factor :/

  2. Babynini
    • one year ago
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    (2sin(theta) ? ___)(sin(theta) ? ____)

  3. freckles
    • one year ago
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    testing what you said for my ownself: \[2(1-\sin^2(\theta))+\sin(\theta)-1=0 \\ \text{ by pythagorean identitiy; also subtract 1 on both sides } \\ 2-2\sin^2(\theta)+\sin(\theta)-1=0 \\ \text{ by distribute property } \\ -2\sin^2(\theta)+\sin(\theta)+1=0 \text{ combining like terms }\] Think you left a sign off on the first term here. I'm also going to multiply -1 on both sides: \[2 \sin^2(\theta)-\sin(\theta)-1=0\]

  4. Babynini
    • one year ago
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    ah yeah I didn't notice the 2sin^2theta was negative.

  5. freckles
    • one year ago
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    anyways if you are doing trial factors you don't have a lot of trials here since -1 is 1(-1) or -1(1)

  6. Babynini
    • one year ago
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    wait so now what do i do for factoring =.=

  7. freckles
    • one year ago
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    if you don't like factoring much you could use the quadratic formula

  8. Babynini
    • one year ago
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    (2sin(theta) - 1)(sin(theta)+1) = 0 ?

  9. Babynini
    • one year ago
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    I think factoring is usually nicer xD

  10. Babynini
    • one year ago
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    dangit. that wouldn't give us the right answer.

  11. freckles
    • one year ago
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    well if you want to do factoring if you multiply that out the middle term will be 2sin(theta)-sin(theta which will not be -sin(theta)

  12. freckles
    • one year ago
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    so which where the -1 and 1 are

  13. Babynini
    • one year ago
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    right.

  14. freckles
    • one year ago
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    switch *

  15. Babynini
    • one year ago
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    huh? wouldn't that give the same answer?

  16. freckles
    • one year ago
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    \[2 \sin^2(\theta)-\sin(\theta)-1=0 \\ (2 \sin(\theta)+1)(\sin(\theta)-1)=0\]

  17. Babynini
    • one year ago
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    which would be 2sin^2(theta)-2sin(theta)+sin(theta)-1=0 which is what we want yeah?

  18. freckles
    • one year ago
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    yep

  19. Babynini
    • one year ago
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    I was confused about -2sin(theta) - sin(theta) = -sin(theta) but now I see hah

  20. freckles
    • one year ago
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    no no -2sin(theta)+sin(theta)=-sin(theta)

  21. freckles
    • one year ago
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    oh did you mean to write that

  22. Babynini
    • one year ago
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    Shiz yeah, sorry.

  23. Babynini
    • one year ago
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    Finals week man. Killing me.

  24. Babynini
    • one year ago
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    K so now we have sin(theta)=1/2 and sin(theta)=1

  25. freckles
    • one year ago
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    almost

  26. Babynini
    • one year ago
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    -1/2?

  27. freckles
    • one year ago
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    \[2 \sin^2(\theta)-\sin(\theta)-1=0 \\ (2 \sin(\theta)+1)(\sin(\theta)-1)=0 \\ \sin(\theta)=\frac{-1}{2} \text{ or } \sin(\theta)=1 \] last equation right just a sign off on the first

  28. Babynini
    • one year ago
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    K and sin = -1/2 at 5pi/6 and 7pi/6 ....right

  29. freckles
    • one year ago
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    can I ask if you want me to check your answers can you give me the intervals in which you want to solve them

  30. Babynini
    • one year ago
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    hm? it just says solve the given equation. No interval.

  31. freckles
    • one year ago
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    probably looking for all solutions then

  32. Babynini
    • one year ago
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    so perhaps 5pi/6, 7pi/6, pi/2

  33. freckles
    • one year ago
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    ok but sin(5pi/6)=1/2 not -1/2

  34. Babynini
    • one year ago
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    crap. 11pi/6

  35. freckles
    • one year ago
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    sin(7pi/6)=-1/2 so 7pi/6 is a solution sin(11pi/6)=-1/2 so 11pi/6 is a solution yep yep

  36. freckles
    • one year ago
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    and yes sin(pi/2)=1 so pi/2 is a solution to the equation sin(u)=1

  37. freckles
    • one year ago
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    but

  38. freckles
    • one year ago
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    if they want all the solutions and since we are working with sin and cos just +2pi*n and say where n is an integer

  39. freckles
    • one year ago
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    \[\theta=\frac{7\pi}{6}+2 \pi n \\ \theta=\frac{11\pi}{6}+2 \pi n \\ \theta=\frac{\pi}{2}+2 \pi n \\ \text{ where } n \text{ is integer }\]

  40. Babynini
    • one year ago
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    Ah ok. I remember this :)

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