Solving Equations

ninatemple

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Jun 25, 2015
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  • 3sin^2(x) - 2cos(x) - 3 = 0, for 0 < x < 180.
    3(1-cos^2(x)) - 2cos(x) - 3 = 0
    3 - 3cos^2(x) 2cos(x) - 3 = 0
    -3cos^2(x) - 2cos(x) = 0
    -3cos(x) = 0 or cos(x) = 0
    (error) x = cos^-1 0
    x = 90


    Therefore, x = 90, 360 - 90
    x = 90, 270


    But the correct answer is 90, 131.8
 
  • 3sin^2(x) - 2cos(x) - 3 = 0, for 0 < x < 180.
    3(1-cos^2(x)) - 2cos(x) - 3 = 0
    3 - 3cos^2(x) - 2cos(x) - 3 = 0
    -3cos^2(x) - 2cos(x) = 0

    cos(x) * [3cos(x) + 2] =0 ..... edit .... changed "-" to "+"

    Two solutions for cos(x)

    cos(x) = 0 → x = 0 ± n * π → x = 90 (for the given domain 0 < x < 180)

    OR

    3cos(x) + 2 = 0 → cos(x) = -2/3 → continue.......
    ..... edit .... changed "-" to "+"



    -3cos(x) = 0 or cos(x) = 0
    (error) x = cos^-1 0
    x = 90


    Therefore, x = 90, 360 - 90
    x = 90, 270


    But the correct answer is 90, 131.8
.
 
Last edited by a moderator:
3sin^2(x) - 2cos(x) - 3 = 0, for 0 < x < 180.


3(1-cos^2(x)) - 2cos(x) - 3 = 0
3 - 3cos^2(x) 2cos(x) - 3 = 0
What happened to the "minus" sign in front of the "2cos(x)"?

-3cos^2(x) - 2cos(x) = 0
Where did the "minus" sign come from? (This sort of algebra "magic" leads to serious errors once one arrives at more-complicated contexts, such as trigonometry. It can also lead to zero points for any work past the "magic" step.)

-3cos(x) = 0 or cos(x) = 0
The last line above implies that you factored the quadratic form -3Y2 - 2Y to get (-3Y)(Y). How did this "factoring" happen? In particular, where went the subtraction? :shock:

Instead, try using what you learned back in algebra. Start with the equation:

. . . . .\(\displaystyle -3Y^2\, -\, 2Y\, =\, 0\)

Multiply through by -1 (or, which results in the same thing, add the terms to the other side of the equation) to get nicer signs:

. . . . .\(\displaystyle 3Y^2\, +\, 2Y\, =\, 0\)

Then do the simple factoring:

. . . . .\(\displaystyle Y(3Y\, +\, 2)\, =\, 0\)

Then set each factor (rather than what appears to be each term) equal to zero, and solve. ;)
 
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