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Sona

  • 2 years ago

please help 2cos(^2)2x=-1+3cos2x

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  1. ByteMe
    • 2 years ago
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    let \(\large y=cos(2x) \) so that trig equation can be written as: \(\large 2y^2=-1+3y \) or \(\large 2y^2-3y+1=0 \) can you solve for y?

  2. Sona
    • 2 years ago
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    so i plug the value for y into one of the equations?

  3. ByteMe
    • 2 years ago
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    you realize that the trig equation you have is actually a trig equation in quadratic form??? so you would solve the quadratic equation by factoring: \(\large 2y^2-3y+1=0 \) \(\large (2y-1)(y-1)=0 \) so y = ????

  4. Sona
    • 2 years ago
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    but its \[2\cos ^{2}2\theta=-1+3\cos 2\theta\]

  5. Sona
    • 2 years ago
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    n dik wat 2 do w/the 2 in front of the theta

  6. ByteMe
    • 2 years ago
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    read what i said... the trigonometric equation you have is a trig equation in QUADRATIC form.... to solve that trig equation, you would use the technique used to solve the quadratic equation: \(\large 2y^2=-1+3y \)

  7. Sona
    • 2 years ago
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    so if i use the quadratic formula then i shud b able to solve for y?

  8. ByteMe
    • 2 years ago
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    ok... this is your equation: \(\large 2\cdot \color {red}{cos^22\theta}=-1+3\cdot \color {red}{cos2\theta} \) \(\large 2\cdot \color {red}{y^2}=-1+3\cdot \color {red}{y} \) sure you can use the quadratic formula but it would be easier to factor as i did earlier...

  9. Sona
    • 2 years ago
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    is y \[(-3\pm sort{17})/4\]

  10. ByteMe
    • 2 years ago
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    no... rewrite the quadratic in standard form, then factor: \(\large 2y^2=-1+3y \) \(\large 2y^2-3y+1=0 \) standard form \(\large (2y-1)(y-1)=0 \) factor so \(\large 2y-1=0 \) or \(\large y-1=0 \) now it's obvious y=1/2 , y = 1....

  11. Sona
    • 2 years ago
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    n y stands for cos2theta?

  12. ByteMe
    • 2 years ago
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    now that you have y=1/2 or y=1, remember that we replaced cos22θ=y so now you have to solve these two equations: \(\large cos2\theta=\frac{1}{2} \) ; \(\large cos2\theta=1 \)

  13. ByteMe
    • 2 years ago
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    yes... y=cos2theta...

  14. Sona
    • 2 years ago
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    so its 30

  15. Sona
    • 2 years ago
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    n 0

  16. ByteMe
    • 2 years ago
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    you have several answers between [0, 2pi)

  17. Sona
    • 2 years ago
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    so 30 150 210 n 330?

  18. Sona
    • 2 years ago
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    or just is it just 30 n 330 since we are dealing with cos?

  19. ByteMe
    • 2 years ago
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    i only see 30, 330, 0 degrees

  20. ByteMe
    • 2 years ago
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    well, the 30 and 330 degrees came from the first equation: \(\large cos2\theta=\frac{1}{2} \) and the 0 degrees came from cos2theta = 1

  21. Sona
    • 2 years ago
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    thank u!!!!!!! u helped so much.....by any chance cud u help me solve this next one? 6tanθ−6cotθ=0

  22. ByteMe
    • 2 years ago
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    yw... post it up as a new question...

  23. Sona
    • 2 years ago
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    okie dokie :)

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