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anonymous

  • one year ago

Find all solutions in the interval [0, 2π). 2 sin2x = sin x

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  1. Nnesha
    • one year ago
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    set it equal to 0

  2. Nnesha
    • one year ago
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    let sin =x to make it easy :D \[\huge\rm 2 x^2 = x\]set it equal to 0 and then take out common factor

  3. anonymous
    • one year ago
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    Can you elaborate more?

  4. Nnesha
    • one year ago
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    |dw:1437488293143:dw| alright so first step ) subtract x b oth sides

  5. Nnesha
    • one year ago
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    let me know what yo uget

  6. anonymous
    • one year ago
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    x^2 = 0

  7. Nnesha
    • one year ago
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    no

  8. Nnesha
    • one year ago
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    |dw:1437488395098:dw| -x stay same at left side

  9. Nnesha
    • one year ago
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    now factor that equation

  10. anonymous
    • one year ago
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    X(2^2-1) = 0 X(4-1) = 0 X(3)=0

  11. Nnesha
    • one year ago
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    hmm it's x^2 not 2 to the 2 power

  12. Nnesha
    • one year ago
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    but yeah x is common factor so take it out \[\huge\rm x(2x-1)\] now set x and (2x-1) both equal to zero then solve for x

  13. anonymous
    • one year ago
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    x= 1/2 so i would have to look at the unit circle to find the ones with 1/2 as the sine value between 0 and 2π or 0 and 180

  14. Nnesha
    • one year ago
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    not between 0 -180 0 -360

  15. Nnesha
    • one year ago
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    at what point sin x = 1/2 and sin x = 0 too

  16. anonymous
    • one year ago
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    Yes. thank you

  17. Nnesha
    • one year ago
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    my pleasure :=)

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