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DanielaJohana

  • one year ago

solve the equation 2sin^2 (x)cos(x)=cos(x) on the interval [0,2pi)

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  1. DanielaJohana
    • one year ago
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    \[2\sin ^{2}(x)\cos(x)=\cos(x)\]

  2. Nnesha
    • one year ago
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    Set it equal to zero and then take out the common factor

  3. DanielaJohana
    • one year ago
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    \[2\sin ^{2}(x)\cos(x)-\cos(x)=0\] like this?

  4. Nnesha
    • one year ago
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    yes right :)

  5. Nnesha
    • one year ago
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    so what is the common factor ?

  6. DanielaJohana
    • one year ago
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    At the beginning, couldn't I have just divided both sides by cos(x)? that would have gotten rid of it on both sides and have set it equal to zero @Nnesha

  7. Nnesha
    • one year ago
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    right!!! i don't why i was thinking that there is a sign between cos and sin you're right!

  8. Nnesha
    • one year ago
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    \[2\sin^2=0\] now solve for sin

  9. Nnesha
    • one year ago
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    wait hmm

  10. Nnesha
    • one year ago
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    hmm i don't think so... cuz if we divide both sides by cos(x)we will get just sin function

  11. Nnesha
    • one year ago
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    i would not divide by cos.

  12. Nnesha
    • one year ago
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    \[\huge\rm 2\sin^2(x)\cos(x)=\cos(x)\] is this your question ?

  13. DanielaJohana
    • one year ago
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    @Nnesha yes

  14. Nnesha
    • one year ago
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    yeah okay let's say sin=x and cos = y \[\huge\rm 2x^2y=y\] you would subtract y both sides not divide

  15. DanielaJohana
    • one year ago
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    yes, but since the question is asking to solve the equation, isn't it asking for when the equations intercept?

  16. Nnesha
    • one year ago
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    ye they are asking for solutions (x,y) <-- exact value of cos and sin between 0 to 2pi

  17. Nnesha
    • one year ago
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    \[\huge\rm \frac{ 2\sin^2(x)\cancel{\cos(x) }}{ \cancel{\cos(x}) } = \frac{ \cancel{\cos(x) }}{ \cancel{\cos(x)}}\] \[2\sin^2(x)=1\] but if you subtract cos(x) both sides you will get \[\large\rm \color{Red}{2\sin^2\cos(x)- \cos(x)=0}\]\[\cos(x)[2\sin^2(x)]=0\] so after that you should set both function equal to zero in this way you will get cos and sin both function value

  18. Nnesha
    • one year ago
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    well you confused me lol let me confirm

  19. Nnesha
    • one year ago
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    btw i'll go with 2nd method 1) subtract 2) find common factor :)

  20. Nnesha
    • one year ago
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    o.O

  21. DanielaJohana
    • one year ago
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    okay I understood how to do it from the comment before it was deleted

  22. freckles
    • one year ago
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    we can't divide both sides by 0 and cos(x) can be 0 -- but yeah you will end up with two equations to solve in the end: cos(x)=0 and 2 sin^2(x)=1

  23. Nnesha
    • one year ago
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    ahh freckles you here i was looking for u but you wer offline ;~;

  24. freckles
    • one year ago
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    \[2 \sin^2(x)\cos(x)-\cos(x)=0 \\ \cos(x)(2\sin^2(x)-1)=0 \\ \cos(x)=0 \text{ or } 2 \sin^2(x)-1=0 \\ \cos(x)=0 \text{ or } 2\sin^2(x)=1\]

  25. Nnesha
    • one year ago
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    forgot 1 there facepalm

  26. DanielaJohana
    • one year ago
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    okay, so for cos(x)=0 that would be pi/2 and 3pi/2 correct?

  27. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @Nnesha \[\huge\rm \frac{ 2\sin^2(x)\cancel{\cos(x) }}{ \cancel{\cos(x}) } = \frac{ \cancel{\cos(x) }}{ \cancel{\cos(x)}}\] \[2\sin^2(x)=1\] but if you subtract cos(x) both sides you will get \[\large\rm \color{Red}{2\sin^2\cos(x)- \cos(x)=0}\]\[\cos(x)[2\sin^2(x)]=0\] so after that you should set both function equal to zero in this way you will get cos and sin both function value \(\color{blue}{\text{End of Quote}}\) correction \[\large\rm \color{Red}{2\sin^2\cos(x)- \cos(x)=0}\]\[\cos(x)[2\sin^2(x)-1]=0\]

  28. Nnesha
    • one year ago
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    ye that's right

  29. DanielaJohana
    • one year ago
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    yes, so the other one would be \[\sin ^{2}(x)=\frac{ 1 }{ 2 }\]

  30. Nnesha
    • one year ago
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    yep but that's sin^2 so take square root both sides

  31. DanielaJohana
    • one year ago
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    \[\sin(x)=\pm \frac{ \sqrt{2} }{ 2 }\] ?

  32. Nnesha
    • one year ago
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    yes right

  33. DanielaJohana
    • one year ago
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    okay so that's pi/4, 3pi/4, 5pi/4, and 7pi/4

  34. Nnesha
    • one year ago
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    nice

  35. DanielaJohana
    • one year ago
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    I get it thank you!

  36. Nnesha
    • one year ago
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    np :)

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