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shaqadry

  • 3 years ago

how to solve 3 sin^2 2x + 2 cos 2x - 2 = 0

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  1. Azteck
    • 3 years ago
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    \[3\sin^{2}(2x)= 3(1-\cos^{2}(2x))\] \[=3-3\cos^2(2x)\] ---------------------------------------------------------------------- \[3-3\cos^2(2x)+2cos(2x)-2=0\] Collect like terms and move everything to the RHS \[3\cos^2(2x)-2cos(2x)-1=0\] Let u=cos(2x) \[3u^2-2u-1=0\] Solve the quadratic.

  2. shaqadry
    • 3 years ago
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    i dont get it, what happen to sin^2 2x?

  3. shaqadry
    • 3 years ago
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    oh wait just got it!

  4. Azteck
    • 3 years ago
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    Yeah I just made sin^2(2x) in terms of cos(2x).

  5. Azteck
    • 3 years ago
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    So could you solve that quadratic and then after you have the values of u, bring back the cos(2x) by replacing u with cos(2x).

  6. Azteck
    • 3 years ago
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    Then after you've done that, solve normally as if it were cos(x) and then after you get the angles, divide by 2 because you're finding x, not 2x.

  7. Azteck
    • 3 years ago
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    If you're unclear about what you're supposed to do, let me know.

  8. shaqadry
    • 3 years ago
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    okay thanks! i really didnt know how to solve that part bcs it has 2x

  9. shaqadry
    • 3 years ago
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    wait i'll try and calculate the angle

  10. shaqadry
    • 3 years ago
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    the angles i got are 0, 54.75, 125.25, 180, 234.75, 305. 25 and 360

  11. Azteck
    • 3 years ago
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    Awesome work. That's pretty much all of them. I think you need to work on your rounding up. Because for some of the angles, I got either 0.01 more or less than your angles.

  12. Azteck
    • 3 years ago
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    Ah I see, you rounded up before dividing by 2. That's okay. You're still correct. I just prefer to round up at the end result.

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