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Math Help - General Solutions

  1. #1
    Junior Member Misa-Campo's Avatar
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    General Solutions

    I have no idea about general solutions since i have never seen them, What are all the general solutions that i must memorize to answer all questions?

    and how do i apply these general solutions in this question:

    Find all values of x for which (cosx)^2=sinxcosx

    thanks
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  2. #2
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    mr fantastic's Avatar
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    Quote Originally Posted by Misa-Campo View Post
    I have no idea about general solutions since i have never seen them, What are all the general solutions that i must memorize to answer all questions?

    and how do i apply these general solutions in this question:

    Find all values of x for which (cosx)^2=sinxcosx

    thanks
    \cos^2 x - \sin x \cos x = 0 \Rightarrow \cos x (\cos x - \sin x) = 0.

    Case 1: \cos x = 0

     \Rightarrow x = \frac{(2n+1) \pi}{2} where n is an integer, that is, x = \pm \frac{\pi}{2}, ~ \pm \frac{3\pi}{2}, ~ \pm \frac{5\pi}{2}, ~ ....


    Case 2: \cos x - \sin x = 0 \Rightarrow \sin x = \cos x \Rightarrow \tan x = 1

     \Rightarrow x = \frac{\pi}{4} + n \pi = \frac{(4n + 1) \pi }{4} where n is an integer, that is, x = \frac{\pi}{4}, ~ \frac{5\pi}{4}, ~ \, ...., ~ - \frac{3 \pi}{4}, ~ \frac{-7 \pi}{4}, ~ ....


    It's the appearance of the n that makes the solution general.
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  3. #3
    MHF Contributor
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    Quote Originally Posted by Misa-Campo View Post
    I have no idea about general solutions since i have never seen them, What are all the general solutions that i must memorize to answer all questions?

    and how do i apply these general solutions in this question:

    Find all values of x for which (cosx)^2=sinxcosx

    thanks
    General solutions? Let us find first the solutions.

    (cosX)^2 = sinXcosX
    (cosX)^2 -sinXcosX = 0
    cosX(cosX -sinX) = 0

    cosX = 0
    X = arccos(0) = pi/2, and/or 3pi/2 -------**

    cosX -sinX = 0
    cosX = sinX
    Divide both sides by cosX,
    1 = tanX ------------positive tan value, so X is in the 1st and 3rd quadrants.
    X = arctan(1)
    X = pi/4, and/or (pi +pi/4) = 5pi/4 .................***

    Those solutions, X = pi/2, 3pi/2, pi/4, 5pi/4 are slutions in one revolution only, or from 0 to 2pi only.

    Maybe by general solutions it means for all the possible values of X.
    If so, then,

    a) when cosX = 0,
    the period of the basic cosine curve y = cosX is 2pi. So the X = pi/2 and 3pi/2 repeats every 2pi. Hence, the general solutions when cosX = 0 are
    X = pi/2 +n(2pi)
    and, X = 3pi/2 +n(2pi)
    where n is any integer.

    b) when cosX -sinX = 0, or when tanX = 1,
    the period od the basic y = tanX curve is pi only.
    Hence the general solutions when tanX = 1 are
    X = pi/4 +n(pi)
    and X = 5pi/4 +n(pi)

    But when n=1, pi/4 +1(pi) = 5pi/4 .....which is 5pi/4 +0(pi).
    And 5pi/4 +1(pi) = 9pi/4 .....which is pi/4 +2(pi).
    Meaning they overlap or are the same angles.

    So, for tanX = 1, the general solution is X = pi/4 +n(pi) only.
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