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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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    in addition to Gerry's solution, note that plotting/sketching the graph is extremely useful for this type of questions!2011-04-13

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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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    @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.2011-05-05