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I recently got this question only half correct:

"Solve for values of $\theta$ the equation $5\cos\theta = 3\cot\theta$, in the interval $0 \leq \theta \leq 360$"

My solution was:

$$5 \cos\theta = 3 \cot\theta$$ $$\frac{\cos\theta}{\cot\theta} = \frac{3}{5}$$ $$\frac{\cos\theta}{\frac{\cos\theta}{\sin\theta}} = \frac{3}{5}$$ $$\frac{\sin\theta \cos\theta}{\cos\theta} = \frac{3}{5}$$ $$\sin\theta = \frac{3}{5}$$ $$\theta = 36.9^\circ, 143^\circ (3 s.f.)$$

Their solutions were the above two angles but also the solutions from $\cos\theta = \frac{3}{5}$ which were 90 & 270. The textbook says "Do not cancel $\cos\theta$ on each side, multiply through by $\sin\theta$" but they do not explain why.

I understand how they get the extra two solutions after taking their approach, but I do not understand why I must take their approach, since I can get rid of the $\cos\theta$.

Any tips would be much appreciated, thanks!

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    $\begingroup$ in addition to Gerry's solution, note that plotting/sketching the graph is extremely useful for this type of questions! $\endgroup$
    – picakhu
    Commented Apr 13, 2011 at 14:26

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Your very first step was dividing both sides by the cotangent. That's a no-no if said cotangent is zero. That's where you lost some solutions.

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    $\begingroup$ Ah I see thank you. so if the interval instead guaranteed the cotangent would never be 0, it would be OK to use my method, but otherwise not, have I understood correctly? $\endgroup$
    – Danny King
    Commented Apr 13, 2011 at 13:53
  • $\begingroup$ @Danny: There is no way of knowing that it guarantees that. But if you were explicitly told that, then you are right. $\endgroup$
    – picakhu
    Commented Apr 13, 2011 at 14:25
  • $\begingroup$ @Danny, yes: the original equation is logically equivalent to "$\sin\theta=3/5{\rm\ OR\ }\cot\theta=0$," so if you're in an interval where you know for a fact that the cotangent isn't zero, your method is fine. $\endgroup$ Commented Apr 14, 2011 at 4:55
  • $\begingroup$ Belated question: If one can´t divide by an unknown that could be 0, why is it acceptable to e.g. simplify $x^2 = x$ to $x = 1$, which is dividing by $x$, but $x$ belongs to the set of real numbers and hence could be 0? (Assuming x does belong to the reals, as is usual in the basic alebgra exercises I´m considering) $\endgroup$
    – Danny King
    Commented May 5, 2011 at 10:22
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    $\begingroup$ @Danny, it is not acceptable to simplify $x^2=x$ to $x=1$. This "simplification" throws away a solution, namely, the solution $x=0$. The only way the simplification might be acceptable is if the equation is coming from some word problem in which it is clear that $x=0$ is not going to be a solution. $\endgroup$ Commented May 5, 2011 at 12:55

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