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zed963
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In this question, why is it possible to cancel out cos theta, doesn't this miss a set of solutions?
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brianeverit
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(Original post by zed963)


In this question, why is it possible to cancel out cos theta, doesn't this miss a set of solutions?
because it is excluding the case when \cos\theta=0
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zed963
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(Original post by brianeverit)
because it is excluding the case when \cos\theta=0
Why's that?
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TeeEm
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(Original post by zed963)
Why's that?
as the previous poster explained you will lose solutions

you can divide only by quantities that cannot equal zero or if they are zero you simply do not care about these lost solutions.

as an example

4x2-5x=0

do you divide this equation by x?
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zed963
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(Original post by TeeEm)
as the previous poster explained you will lose solutions

you can divide only by quantities that cannot equal zero or if they are zero you simply do not care about these lost solutions.

as an example

4x2-5x=0

do you divide this equation by x?
You could if you wanted to, or you could factorise it.
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TeeEm
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(Original post by zed963)
You could if you wanted to, or you could factorise it.
you cannot divide by x because you will lose the solution x=0 (unless x=0 is of no interest in the problem you are solving)

only factorize.

same with your trig

otherwise you will lose the solutions from cosx =0

hope it helps
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davros
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(Original post by zed963)


In this question, why is it possible to cancel out cos theta, doesn't this miss a set of solutions?
You're not really "cancelling" cos theta in the same way as when you divide both sides of an equation by x - the point is that at those points where cos theta = 0, both cos theta and tan theta are undefined - they're not valid solutions of the original equation. Therefore it is OK to multiply the original equation by cos theta to simplify it.
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