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!

  • 1
    $\begingroup$ in addition to Gerry's solution, note that plotting/sketching the graph is extremely useful for this type of questions! $\endgroup$ – picakhu Apr 13 '11 at 14:26

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.

| cite | improve this answer | |
  • 1
    $\begingroup$ 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 Apr 13 '11 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 Apr 13 '11 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. $\endgroup$ – Gerry Myerson Apr 14 '11 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 May 5 '11 at 10:22
  • 1
    $\begingroup$ @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. $\endgroup$ – Gerry Myerson May 5 '11 at 12:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.