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    I'm a bit confused by what my teacher was saying about why you can't divide by sin x or cos x etc? so if x was zero i assume you would miss a solution, but couldn't you just find this by sketching a graph at the end? thanks
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    (Original post by Bertybassett)
    I'm a bit confused by what my teacher was saying about why you can't divide by sin x or cos x etc? so if x was zero i assume you would miss a solution, but couldn't you just find this by sketching a graph at the end? thanks
    In what context?

    If you are solving a trigonometric equation like \sin x \cos(x) = \frac{1}{2} \sin x between, say 0 \leq  x \leq 360, then no you cannot divide by sin(x) because there exist at least one x in the stated interval for which \sin(x) = 0, and so you would be dividing by 0 which is not allowed.

    This is why you need to factor out instead, and say that the solutions are given by \sin x = 0 or \cos x = \frac{1}{2} then solve the two equations separately for the values of x. Division by \sin x would force you to completely disregard solutions to \sin x = 0 hence miss out on a few valid solutions to the problem.

    You can sketch a graph, but quite often as indicated here, the graph is not obvious and hence it's must easier to tackle the problem analytically rather than graphically.
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    (Original post by RDKGames)
    In what context?

    If you are solving a trigonometric equation like \sin x \cos(x) = \frac{1}{2} \sin x between, say 0 \leq  x \leq 360, then no you cannot divide by sin(x) because there exist at least one x in the stated interval for which \sin(x) = 0, and so you would be dividing by 0 which is not allowed.

    This is why you need to factor out instead, and say that the solutions are given by \sin x = 0 or \cos x = \frac{1}{2} then solve the two equations separately for the values of x. Division by \sin x would force you to completely disregard solutions to \sin x = 0 hence miss out on a few valid solutions to the problem.

    You can sketch a graph, but quite often as indicated here, the graph is not obvious and hence it's must easier to tackle the problem analytically rather than graphically.
    Thanks for the reply, so when you say in what context, do you mean that it is usually unacceptable to divide by sinx or cos x etc, and should this never be done in an exam?

    Also, for your example of sinxcosx=0.5sinx, i tried solving this and so took away the 0.5sinx from both sides, then factorised sinx out of it, and got solutions of 0, 180, 360, 60, 300. have i done this correctly? thanks
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    (Original post by Bertybassett)
    Thanks for the reply, so when you say in what context, do you mean that it is usually unacceptable to divide by sinx or cos x etc, and should this never be done in an exam?
    I just mean what sort of topic you're talking about. But generally, the rule is that you cannot divide by an expression unless you are certain that it cannot equal 0.

    So if I have 2x=x^2 for x>0 as an example, then obviously x is a common factor. Can it be 0? No because x>0 does not include 0, so I can divide by it and find that x=2. If x \geq 0 instead, then dividing would miss a crucial solution of x=0 since this is an acceptable value.

    Also, for your example of sinxcosx=0.5sinx, i tried solving this and so took away the 0.5sinx from both sides, then factorised sinx out of it, and got solutions of 0, 180, 360, 60, 300. have i done this correctly? thanks
    Those are indeed all of the solutions in the given range.
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    (Original post by RDKGames)
    I just mean what sort of topic you're talking about. But generally, the rule is that you cannot divide by an expression unless you are certain that it cannot equal 0.

    So if I have 2x=x^2 for x>0 as an example, then obviously x is a common factor. Can it be 0? No because x>0 does not include 0, so I can divide by it and find that x=2. If x \geq 0 instead, then dividing would miss a crucial solution of x=0 since this is an acceptable value.



    Those are indeed all of the solutions in the given range.
    i just came across a question: solve for 0 to 360 5sin3x = 2cos2x - to solve this i guess first you could say 2x = y, and then divide both sides by 2cosy, but then surely that breaks the rule of how you cant divide by cos of an angle? what would i do in this case? thanks
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    (Original post by MR1999)
    If you want to solve the equation, you need to bring both equations to one side to get

    5sin3x-2cos2x=0.

    Then use the R formula (or Harmonic Addition Theorem) found the in the formulae booklet to combine the functions.
    That's not the approach - the arguments are different.

    (Original post by Bertybassett)
    i just came across a question: solve for 0 to 360 5sin3x = 2cos2x - to solve this i guess first you could say 2x = y, and then divide both sides by 2cosy, but then surely that breaks the rule of how you cant divide by cos of an angle? what would i do in this case? thanks
    What do you mean divide both sides by 2\cos y? Firstly, \cos 2x is indeed 0 in that range, so you cannot divide by it. Secondly, it is not a common factor in the first place for division to be considered.
 
 
 
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