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anonymous

  • 4 years ago

6cos²x-sinx-4=0 x=?

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  1. anonymous
    • 4 years ago
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    i've changed cos²x into 1-sin²x

  2. anonymous
    • 4 years ago
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    6cos²x-sinx-4=0 6(1-sin²x)-sinx-4=0 -6sin²x-sinx+2=0 -(6sin²x+sinx-2)=0 then i've used quadratic formula. a=6 b=1 c=-2

  3. anonymous
    • 4 years ago
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    then i got \[\sin x = (-1±\sqrt{1+48})/12\]

  4. anonymous
    • 4 years ago
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    which means sinx=2/3 or sinx=-1/2 when i solved for each of them, didn't give me the correct answer. what did i do wrong?

  5. lalaly
    • 4 years ago
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    you can let u=sinx -(6u^2+u-2)=0 -(2u-1)(3u+2)=0

  6. lalaly
    • 4 years ago
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    u=-1/2 u=2/3 so sinx=-1/2 sinx=2/3

  7. anonymous
    • 4 years ago
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    yep then sin is positive in QI and QII and sin is negative in QIII and QIV

  8. anonymous
    • 4 years ago
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    so for sinx=2/3, the r.a.a is sin^-1(2/3) = 0.73

  9. anonymous
    • 4 years ago
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    which means x = 0.73 or 2.41 since they exist in QI and QII

  10. anonymous
    • 4 years ago
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    and for sinx=-1/2, the r.a.a is sin^-1(1/2) = π/6 and they exist in QIII and QIV so x = 7π/6 and 11π/6

  11. anonymous
    • 4 years ago
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    i got final answer as x=0.73, 2.41, 7π/6, and 11π/6 but they're not correct. anything i did wrong there?

  12. asnaseer
    • 4 years ago
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    your solution is correct - why do you say its wrong?

  13. anonymous
    • 4 years ago
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    the answer sheet says x = π/6, 5π/6, 3.87, 5.55 so I've tried one of my answer, 7π/6 and subbed it in to the equation, and it gave me 1 instead of 0, meaning that my answers are incorrect. :/

  14. asnaseer
    • 4 years ago
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    inverse sin of 0.5 is \(\pi/6\) and \(5\pi/6\)

  15. anonymous
    • 4 years ago
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    inverse sin of 0.5 is π/6, but since sinx = -1/2, they should exist in QIII or QIV meaning that x=7π/6 or 11π/6..

  16. anonymous
    • 4 years ago
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    sin is negative in QIII or QIV

  17. asnaseer
    • 4 years ago
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    hang on - let me recheck your steps...

  18. anonymous
    • 4 years ago
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    ok thanks for the help!

  19. asnaseer
    • 4 years ago
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    ah - ok - I see the mistake

  20. asnaseer
    • 4 years ago
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    when you substitute \(\cos^2(x)=1-\sin^2(x)\) you made an error in that step. it should have resulted in:\[6\cos^2(x)-\sin(x)-4=0\]\[6(1-\sin^2(x))-\sin(x)-4=0\]\[6-6\sin^2(x)-\sin(x)-4=0\]

  21. anonymous
    • 4 years ago
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    hm? 6-6sin²x-sinx-4=0 -6sin²x-sinx+2=0 -(6sin²x+sinx-2)=0 no? am i still doing something wrong? :/

  22. asnaseer
    • 4 years ago
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    and then:\[-6\sin^2(x)-\sin(x)+2=0\]\[-(2\sin(x)-1)(3\sin(x)+2)=0\]giving you:\[\sin(x)=-2/3\]or:\[\sin(x)=0.5\]

  23. anonymous
    • 4 years ago
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    ohh okay. I have no idea why I got the wrong answer for using quadratic formula for that.. somehow I could not factor that. i have no idea why but thanks for the help!

  24. asnaseer
    • 4 years ago
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    yw

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