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!
Best Answer
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.