# Solving simple trigonometrical equations for multiple solutions

I have to find all angles between $0^o$ and $360^o$ inclusive which satisfy the equation $$4\,sin(y)\,cos(y)=tan(y)$$ Here is what I did \require{cancel}\begin{align}4\,\cancel{sin(y)}\,cos(y)&=\frac {\cancel{sin(y)}}{cos(y)}\\ 4\,cos^2(y)&=1\\ \sqrt{cos^2(y)}&=\sqrt{\frac14}\\ cos(y)&=\pm \frac12\\ y&=60,\,120,\,240,\,300 \end{align} The problem is, that the answer also includes $y=0,\,180,\,360$. How?

When you cancel the $(\sin y)$'s, what you're really doing is dividing both sides by $\sin y$. But what if $\sin y = 0$? Then you just performed an illegal operation. You'll need to treat that as a separate case.

Instead, it's a lot cleaner to solve it by bringing everything to one side and then factoring: \begin{align*} 4\sin y \cos y &= \frac{\sin y}{\cos y} \\ 4\sin y \cos y - \frac{\sin y}{\cos y} &= 0 \\ 4\sin y \cos^2 y - \sin y &= 0 \\ (\sin y)(4\cos^2 y - 1) &= 0 \\ (\sin y)(2\cos y + 1)(2\cos y - 1) &= 0 \end{align*} This yields three cases: $$\begin{cases} \sin y = 0 &\implies& y = 0^\circ, 180^\circ, 360^\circ \\ 2\cos y + 1 = 0 \implies \cos y = \frac{-1}{2} &\implies& y = 120^\circ, 240^\circ \\ 2\cos y - 1 = 0 \implies \cos y = \frac{1}{2} &\implies& y = 60^\circ, 300^\circ \\ \end{cases}$$

Hint

$$(ab=ac\iff b=c)\iff a\ne0$$

You are dividing the equation by $siny$ wich is forbidden if $siny=0$. Therefore you have to consider this case separately.

Hint: Let's try to solve $$x^2=x$$

$$x^2=x \implies x=1$$

We missed the solution $0$ why ? with same reason as in your solution.