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Rosy95

  • 3 years ago

PreCal Help Please

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  1. Rosy95
    • 3 years ago
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    Solve\[4\sin^2x+4\sqrt{2} cosx-6=0\]

  2. amoodarya
    • 3 years ago
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    the question is not here ! or i do not see it

  3. Rosy95
    • 3 years ago
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    I just posted it before you wrote your comment

  4. hba
    • 3 years ago
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    @amoodarya Refresh your page

  5. cgreenwade2000
    • 3 years ago
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    Substitute in for the value of sin^2(x) You know what it is?

  6. Rosy95
    • 3 years ago
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    Substitute it where? I am still confused.

  7. cgreenwade2000
    • 3 years ago
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    sin^2(x) can be put in terms of cos^2(x)

  8. cgreenwade2000
    • 3 years ago
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    sin^2(x) = 1-cos^2 (x) Another really important one to remember. (I hope you're writing these important ones down, especially if you plan on going further in calculus)

  9. hartnn
    • 3 years ago
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    then put cos x =y and you get a Quadratic in 'y', which can be solved easily.

  10. hba
    • 3 years ago
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    ^ this sounds better

  11. cgreenwade2000
    • 3 years ago
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    I wasn't going to suggest that.......but it makes it a lot easier. Go that route.

  12. cgreenwade2000
    • 3 years ago
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    Don't forget to distribute when you substitute for sin^2(x)

  13. Rosy95
    • 3 years ago
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    wait so is the new equation\[4( \frac{ 1 }{\cos^2x }) + 4 \sqrt{2}cosx-6=0\]

  14. cgreenwade2000
    • 3 years ago
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    4(1-cos^2x).......

  15. hartnn
    • 3 years ago
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    actually, \[\huge \color{red}{\sin^2x+\cos^2x=1 \\ \implies \sin^2x =1-\cos^2x}\]

  16. amoodarya
    • 3 years ago
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    no |dw:1358282546908:dw|

  17. hartnn
    • 3 years ago
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    \[\huge \color{red}{\sin^2x+\cos^2x=1 \\ \implies \sin^2x =1-\cos^2x \\ \implies 4\sin^2x=4-4\cos^2x}\]

  18. amoodarya
    • 3 years ago
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    |dw:1358282634437:dw|

  19. amoodarya
    • 3 years ago
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    simplify that

  20. Rosy95
    • 3 years ago
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    Can you rewrite that? It is hard to read

  21. hartnn
    • 3 years ago
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    \[4\sin^2x+4\sqrt{2} cosx-6=0 \\ 4-4\cos^2x+4\sqrt{2} cosx-6=0\\4\cos^2x-4\sqrt{2} cosx+2=0 \\put \cos x=y \\ \implies 4y^2-4\sqrt2y+2=0\] can you solve this quadratic ?

  22. cgreenwade2000
    • 3 years ago
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    That should be a -2 @hartnn

  23. hartnn
    • 3 years ago
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    u sure ?

  24. Rosy95
    • 3 years ago
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    ok so the answer is \[\frac{ \pi }{4 }+2\pi k, \frac{ 7\pi }{4 }+2\pi k\]

  25. hartnn
    • 3 years ago
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    how did u get that ? what values of y u got ?

  26. amoodarya
    • 3 years ago
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    |dw:1358282921662:dw|

  27. cgreenwade2000
    • 3 years ago
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    @hartnn 4-6=-2

  28. hartnn
    • 3 years ago
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    @Rosy95 did you actually solve it ?

  29. amoodarya
    • 3 years ago
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    |dw:1358283161601:dw|

  30. Rosy95
    • 3 years ago
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    Im' sorry, I forgot to include the rest of the question...it is solve (that long equation) for all real values of x.

  31. hartnn
    • 3 years ago
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    @Rosy95 what answer you wrote is correct. did you solve it by yourself ? and any doubts anywhere ?

  32. Rosy95
    • 3 years ago
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    I solved it by myself but i looked at the answer choices when I got half of it.

  33. Rosy95
    • 3 years ago
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    Thanks everyone!

  34. hartnn
    • 3 years ago
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    ok, welcome ^_^

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