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Thread: a sin θ + b cos θ = 0

  1. #1
    Nov 2008

    Exclamation a sin θ + b cos θ = 0

    When an equation of the form a sin θ + b cos θ is equated to 0, is it possible to divide throughout by cos θ, so that the original trigonometric equation is reduced to a trigonometric equation in terms of tan θ? Or would you be losing a solution in doing so?

    Can anyone tell me what's wrong with the solution of the equation below? I know that by dividing throughout by cos θ, I would have got to the answer, but what's wrong with the method below?

    Solve 3^0.5 sin θ - cos θ = 0 for 0<θ<360

    Since sin θ = cos θ tan θ, I worked it out as follows:

    3^0.5 cos θ tan θ - cos θ = 0
    cos θ (3^0.5 tan θ - 1) = 0

    so either cos θ = 0 or 3^0.5 tan θ - 1 = 0
    θ = 90 or θ = 30

    Then by writing the general solutions to both, I found out that the angles θ within the range are θ = 0, 30, 210, 270

    First problem: the book which I am using only gives out θ = 30 and 210 as answers and teacher said there are no mistakes with answers

    Second problem: When I substituted the values of θ =0 and 270 in the original equation, I found out that these are not correct values since 3^0.5 sin 0 - cos 0 is not equal to 0.

    PLEASE HELP!!!!!!
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  2. #2
    Junior Member
    Feb 2009
    3^0.5 sin A=cosA


    or,4 sin^2 A=1

    or,sin A=(+-)1/2


    but,A=150,330 deg. are trivial cases.

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  3. #3
    Nov 2008
    Many thanks for the alternative correct methods, but I have two queries:

    what's the problems with the method I used?
    How have you decided that 150 is a trivial angle?

    Thanks a lot
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  4. #4
    MHF Contributor
    Grandad's Avatar
    Dec 2008
    South Coast of England

    Trig equations

    Hello yobacul and sbcd90

    You have to be a little more careful when you're solving equations that you don't introduce solutions that aren't really there. Both of you have done that in what you have written. In the first case the values $\displaystyle \theta = 0^o$ and $\displaystyle 270^o$ didn't satisfy the original equation, and in the second case $\displaystyle A = 150^o$ and $\displaystyle 330^o$ didn't either.

    So, there are two questions to be answered:

    • What is the correct way of solving this equation?
    • Where have these extra answers come from?

    The answer to the first question is the one that yobacul gave at the beginning: divide throughout by $\displaystyle \cos\theta$, and then solve $\displaystyle \sqrt 3\tan\theta - 1 = 0$.

    To answer the second question (and there are really two questions here, because both alternative methods introduced extra, unwanted, answers) consider the following examples:

    (1) Solve the equation $\displaystyle 2x - 3 = 0$

    Multiply both sides by $\displaystyle x$:

    $\displaystyle x(2x - 3) = 0$

    $\displaystyle \Rightarrow x = 0$ or $\displaystyle 1.5$.

    That's rather silly, isn't it? But that's effectively what yobacul did when you multiplied by $\displaystyle \cos\theta$, and then said that $\displaystyle \cos \theta = 0$ is a possible solution!

    (2) Solve the equation $\displaystyle 5x = 10$

    Square both sides:

    $\displaystyle 25x^2 = 100$

    $\displaystyle \Rightarrow x^2 = 4$

    $\displaystyle \Rightarrow x = \pm 2$

    This time, this is effectively what sbcd90 did, which is why the extra answers of $\displaystyle 150^o$ and $\displaystyle 330^o$ were created. In fact these are the solutions to $\displaystyle \tan\theta = -\frac{1}{\sqrt 3}$, not (as they should be) $\displaystyle \tan\theta = \frac{1}{\sqrt 3}$

    I hope this clears up the mystery.

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  5. #5
    Junior Member
    Feb 2009


    yeah ,some trivial solutions may creep in,& for this reason,on squaring,u get some trivial solutions.

    but they may not satisfy the original it is essential to check whether they satisfy the trigonometric equations.
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