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Math Help - Square of equations - Why some solutions rejected

  1. #1
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    Square of equations - Why some solutions rejected

    Dear all,

    I'm solving 2 sums

    a) 1 - cos x = (sqrt 3) sin x

    By squaring both sides, and using the sin^2x + cos^2x = 1 relation,
    I get 2 cos^2x - cos x - 1=0
    Solving, I get x = 120, 240 (for 2nd, 3rd quadrant) and 0, 360 (for 1st, 4th quadrant)
    May I know why 240 is rejected?


    b) 2 + sin x = 4 cos x
    Similarly, I'm solving down to 17 sin^2x + 4 sinx - 12 = 0
    I get sin x = 0.7307 or -0.966
    Hence, x - 46.9 or 133.1 ; 285 or 255
    Why is 133.1 and 255 rejected?

    Thanks
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  2. #2
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    Re: Square of equations - Why some solutions rejected

    The idea is that f(x)=g(x) implies (f(x))^2=(g(x))^2 for all x, but not vice versa. The correct equivalence is

    \text{for all }x,\, [f(x) = g(x)\text{ or }f(x)=-g(x)]\iff (f(x))^2=(g(x))^2.

    Thus, if you solve (f(x))^2=(g(x))^2 and find that, say,

    (f(x))^2=(g(x))^2\iff x=x_0\text{ or }x=x_1,

    it does not follow that

    f(x_1)=g(x_1)\text{ and }f(x_2)=g(x_2).

    At most you can guarantee that

    [f(x_1)=g(x_1)\text{ or }f(x_1)=-g(x_1)]\text{ and }[f(x_2)=g(x_2)\text{ or }f(x_2)=-g(x_2)].

    Indeed, in your first example

    1 - \cos(240^\circ)=3/2 = -\sqrt{3}\sin(240^\circ).
    Thanks from romsek
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  3. #3
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    Re: Square of equations - Why some solutions rejected

    Hello, Rodimus79!

    ;808717]Dear all,

    I'm solving 2 sums

    a)\;1 - \cos x \:=\: \sqrt 3 \sin x

    By squaring both sides, and using the sin^2x + cos^2x = 1 relation,
    I get 2 cos^2x - cos x - 1=0

    Solving, I get x = 120o, 240o (for 2nd, 3rd quadrant) and 0o, 360o (for 1st, 4th quadrant)

    May I know why 240o is rejected?

    When squaring an equation, we must check our answers.
    Some answers may not check out.


    \text{Check }x = 0^o\!:\;\text{Does }1-\cos 0^o\text{ equal }\sqrt{3}\sin0^o\,?

    . . \begin{Bmatrix}1-\cos0^o &=& 1-1 &=& 0 \\ \sqrt{3}\sin0^o &=& \sqrt{3}(0) &=& 0 \end{Bmatrix}\;\text{ Yes!}


    \text{Check }x = 360^o\!:\;\text{Does }1-\cos360^o\text{ equal }\sqrt{3}\sin360^o\,?

    . . \begin{Bmatrix}1-\cos360^o &=& 1 - 1 &=& 0 \\ \sqrt{3}\sin360^o &=& \sqrt{3}(0) &=& 0 \end{Bmatrix}\;\text{ Yes!}


    \text{Check }x = 120^o\!:\;\text{Does }1 - \cos120^o\text{ equal }\sqrt{3}\sin120^o\,?

    . . \begin{Bmatrix}1-\cos120^o &=& 1 - (\text{-}\frac{1}{2}) &=& \frac{3}{2} \\ \sqrt{3}\sin120^o &=& \sqrt{3}(\frac{\sqrt{3}}{2}) &=& \frac{3}{2} \end{Bmatrix}\;\text{ Yes!}


    \textCheck }x = 240^o\!:\;\text{Does }1-\cos240^o\text{ equal }\sqrt{3}\sin240^o\,?

    . . \begin{Bmatrix}1-\cos240^o &=& 1-(\text{-}\frac{1}{2}) &=& \frac{3}{2} \\ \sqrt{3}\sin240^o &=& \sqrt{3}(\text{-}\frac{\sqrt{3}}{2}) &=& \text{-}\frac{3}{2} \end{Bmatrix}\;\;{\color{blue}No!}
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  4. #4
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    Re: Square of equations - Why some solutions rejected

    Hello again!

    Do you know WHY squaring an equation can introduce "extra" solutions?


    Consider the equation: . 2x \:=\:6

    The answer is, of course: . x \,=\,3


    Suppose we square the equation before solving.

    Then we have: . (2x)^2 \:=\:6^2 \quad\Rightarrow\quad 4x^2 \:=\:36
    This is a quadratic equation; it will have two solutions.

    . . . x^2 \:=\:9 \quad\Rightarrow\quad x \:=\:\pm3


    We find that x = 3 satisfies the equation,
    . . but x = \text{-}3 does not.

    Therefore, x=\text{-}3 is an extraneous solution.
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