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Babybee24

  • 3 years ago

cos^2(x)+cos(x)sin(x)=0 solve for x

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  1. myininaya
    • 3 years ago
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    Try factoring

  2. Babybee24
    • 3 years ago
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    I factored a cos(x) out but then im stuck

  3. myininaya
    • 3 years ago
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    cos(x)( ? + ? ) =0 So what should be in the places of those ?'s once you factor out cos(x)?

  4. Babybee24
    • 3 years ago
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    cos(x)(cosx+sinx)

  5. myininaya
    • 3 years ago
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    Yes. Set both factors equal to 0. cos(x)=0 ; cos(x)+sin(x)=0

  6. myininaya
    • 3 years ago
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    Solve cos(x)=0 for x And find the value x so that cos(x) and sin(x) are opposites (i.e cos(x)=-sin(x))

  7. Babybee24
    • 3 years ago
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    cosx=pie/2 ?

  8. myininaya
    • 3 years ago
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    cos(x)=0 for x=pi/2 cos(x) does not equal pi/2 But one value that satisfies cos(x)=0 is x=pi/2 Are there any other values?

  9. Babybee24
    • 3 years ago
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    x=3pi/2

  10. myininaya
    • 3 years ago
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    Ok and you can also say we have x=pi/2+2npi where n is an integer or x=3pi/2+2n pi where n is an integer

  11. myininaya
    • 3 years ago
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    We also need to find the values of x where cos(x) and sin(x) have opposite values.

  12. myininaya
    • 3 years ago
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    |dw:1357920385873:dw| As you can see they will be opposite in quadrant 2 and 4.

  13. myininaya
    • 3 years ago
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    |dw:1357920430907:dw| These points is where cos(x) and sin(x) are opposite. Now what angle are these points at from the initial ray?

  14. Babybee24
    • 3 years ago
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    135 and 315?

  15. myininaya
    • 3 years ago
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    Correct.

  16. myininaya
    • 3 years ago
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    Now should your answer be in radians or degrees?

  17. Babybee24
    • 3 years ago
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    um radians?

  18. myininaya
    • 3 years ago
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    So 135*pi/180 or 315*pi/180

  19. myininaya
    • 3 years ago
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    Don't forget we can go around the circle infinitely many times from those angles.

  20. myininaya
    • 3 years ago
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    So 135pi/180+2npi , 315pi/180+2npi

  21. myininaya
    • 3 years ago
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    Don't forget the other solutions we wrote above. Good job. :)

  22. Babybee24
    • 3 years ago
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    thanks!

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