# Little Trigonometry misconception

Just want to get this little crease in my trig knowledge, ironed out. So, a few months ago, I made the classic newbie mistake. I had a nasty trig equation which I needed to simplify, in order to solve for theta in a given interval. In doing so, I divided by a trig function and consequently I lost a few solutions. When my teacher noticed what I did, he said never divide something by a trig function instead expand the bracket or subtract. When he told me this instead of asking why this was the case, I just placed it into memory and continued striving to become #1 in the class. If I remember correctly, the trig equation I had to solve was: $$\tan(\theta)=\tan(\theta)(2+3\sin(\theta))$$ $$0\leq\theta\leq360$$

The immediate step I took was to divide both sides by $\tan(\theta)$, resulting in 1. When I try to think about why we lose solutions, the following comes to mind: Well, I know division by 0 is undefined and in effect that's what I could be doing since the domain has several spots where $\tan(\theta)=0$. However, I still feel this doesn't fully justify why we may lose some solutions and retain some. Please forgive any grammar mistakes.

• This should not be limited to trigonometric function only. – Mick Oct 26 '16 at 3:55
• Yeah, bad wording. Same reasoning could be applied to any equation where a function in that equation could = 0 in a given interval. However, am I correct in thinking if the domain was such that $0\leq\theta\leq360, \theta \not = m\pi, m\in(0,1,2)$ then I'd well within my rights to divide by $\tan(\theta)$ @Mick – user376253 Oct 26 '16 at 4:07
• The answer is still NO. Cancellation is just division in disguise. Cancellation is allowed only if we are absolutely sure that we are not dividing by zero. We are loosing one or more roots in the course of cancellation. – Mick Oct 26 '16 at 4:37
• @Mick Got it, thanks. – user376253 Oct 26 '16 at 4:42
• Even if those roots are in-admisable (because of physical requirements), you still have to give all of the roots first and then reject the un-acceptable one. Without doing the above means you don't know the rule of the game. – Mick Oct 26 '16 at 4:48

$$\tan(\theta)=\tan(\theta)(2+3\sin(\theta)) \rightarrow \tan(\theta) (1 - 2 - 3 \sin\theta) = 0 \rightarrow\tan\theta(1+3\sin\theta) = 0$$ $$\tan\theta = 0 \implies \theta = k\pi | k \in \mathbb{Z}\quad\lor\quad \sin\theta = -\frac{1}{3} \implies \theta = \arcsin\left(-\frac{1}{3}\right) + 2k\pi | k \in \mathbb{Z}$$ Eliminating $\tan\theta$ will eliminate $k\pi$ solutions
Think of it like this: For what values do we have $$P = NP$$
If you divide through right away by $P$, you get only $N=1$. But $P=0$ results in $0=0$, which is also a solution.