# Explain the concept behind solving $\sin(x)\cos(x) + \cos(x) = 0$, from Paul's Online Math Notes

$$\sin(x)\cos(x) + \cos(x) = 0$$

You are asked to find all possible solutions. What I immediately did was bring over the $\cos(x)$ term and then divided across by $\cos(x)$ and then proceeded from there:

$$\sin(x)\cos(x) = -\cos(x)$$

$$\sin(x) = -1$$

Now it states in the solution which I looked at after that this is wrong. I know now after reading the solution that this will result in losing solutions in my final answer but I don't understand why. I don't think I'm breaking any rules by doing what I did above so why does it result in losing solutions. Can anyone tell me what it is I'm not understanding about this equation.

Thank you to anyone who offers help in advance!

• Safer to factor: $\cos x(\sin x+1)=0$. A product is zero, so one of the factors must be zero. Two possibilities, either $\cos x=0$ or $\sin x=-1$. The “rule” that you violated, as user2825632 pointed out, was to divide by a potential zero. – Lubin Sep 13 '16 at 17:33
• With the same argument, the equation $x=0$ has no solution, because you can remove the factor $x$, leaving $1=0$. ;-) – egreg Sep 13 '16 at 17:39
• Just a note - this isn't a trigonometry issue. It's an algebra issue. – Joel Sep 13 '16 at 21:10

If $\cos(x) = 0$, then you cannot divide by $\cos(x)$. Therefore, you must separately consider the case where $\cos(x) = 0$ and see if that is a valid solution. In this case, it is. That would be the solution you would be missing from your final answer.

• Ok, so basically what you're saying is that my problem was that I didn't consider the actual value of cos(x) before just deciding to use regular old arithmetic with it? – RonGiant Sep 13 '16 at 17:33
• Yep! You should make sure you that you don't accidentally divide by 0. As a simpler example, consider $3x = x^2$. You could divide by $x$ and say $x=3$, but in the case that $x=0$, that division would not be valid. Then we separately consider, "What if $x=0$?" and plug in $x=0$ and see that $x=0$ is also a valid solution. – user2825632 Sep 13 '16 at 17:35
• When you get a good answer to your question the way to thank the author is to accept it (the check mark) and upvote it (the up arrow). You can upvote more than one answer to a question. – Ethan Bolker Sep 13 '16 at 19:29
• An important warning: in general, there is no guarantee that the value you are about to divide by is actually a solution; it so happens that it is a solution here. What you must do is verify whether it's a solution. Afterward, you assert "assuming $cos(x)$ is not zero (I already know what happens if it were zero)", and then divide and move on. – Euro Micelli Sep 13 '16 at 22:20

You have that $$(\sin x + 1)\cos x = 0.$$ If $\cos x = 0$, you cannot divide by $\cos x$. Using the fact that if $ab = 0$ then either $a$ or $b$ is $0$, we have that $\sin x = -1$ or $\cos x = 0$. If $\sin x = -1$ then $x = 3\pi/2 + 2k\pi$, and if $\cos x = 0$ then $x = \pi/2+\ell\pi$ for integral values of $k$ and $\ell$.

Let me show you a related proof. Let's start with some number $a$ and set $b=a$.

$$a = b$$

Multiply by $a$

$$a^2=ab$$

Subtract $b^2$

$$a^2-b^2 = ab-b^2$$

Factor both sides

$$(a+b)(a-b)=b(a-b)$$

Divide a common factor

$$a+b = b$$

Substitute $b$ for $a$ (since they are equal)

$$b+b=b$$

Divide by $b$

$$1+1=1$$

To help see what went wrong, plug in a specific value: if we assume the variables were $5$, then my steps become $5=5$, $25=25$, $25-25 = 25-25$, $(5+5)(0) = 5(0)$, $5+5=5$, $1+1=1$.

So at one step here I divided by $0$, and that is exactly where my equalities start being wrong. If you divide by zero, things can go very, very wrong. In my case, I concluded that $1+1=1$. So each and every time you divide (or cancel) things while doing algebra, you have to check that it's not zero.

• I like this answer because it really brings out the fact that the problem is in basic equation solving, not trig. – Patricia Shanahan Sep 13 '16 at 22:21

The conflict is due to,(as you find) $$sin(x)=-1$$ for all possible values of x, $cos(x)$ lead to $0$ and hence the previous step done by you is wrong. To avoid this conflict you must take all cases as, $$(sin(x)+1)cos(x)=0$$

• The step would also be wrong if the equation had been $\cos x(\sin x+\frac{1}{2})=0$. – egreg Sep 13 '16 at 17:40
• @egreg you are correct but in your equation $cosx(sinx+\frac{1}{2})$ the two posibilities for x can be found by $cosx=0$ and $sinx+\frac{1}{2}=0$ both are considered as correct but in the equation $cosx(sinx+1)=0$ you can not consider $sinx+1=0$ the only possibility is $cosx=0$ (i give my answer in this contrast but you are absolutely correct) – Sahil Kumar Sep 13 '16 at 17:44
• What I wanted to emphasize is the fact that $\sin x=-1$ implies $\cos x=0$ is completely irrelevant. – egreg Sep 13 '16 at 17:48
• @egreg ok i got it...... – Sahil Kumar Sep 13 '16 at 17:49

I would use this $$(\sin(x)+1)\cdot \cos(x)=0\tag{1}$$ and then split the big problem into two small problems - since $a \cdot b = 0$ is "special". $$...=0\quad \text{or}\quad ...=0\tag{2}$$ (If one of these parts of the equation is $0$ then the complete equation will be $0$. Fill the blanks yourself.) After a total of $3$ changes to the equations you will have $$x=...\quad \text{or}\quad x=...\tag{3}$$ You might need to use books/tables to solve the equations to get two sets of solutions. (Remember that sin and cos repeat every $2\pi$ !) After that you can merge these two sets into one set (and eliminate impossible solutions if needed).

[I guess you can write the solution as $x' + k\cdot\pi$. But I'm not sure about that. So you should verify that yourself.]

• I'm not clear what is going on in (2) or (3), what equals $0$ in (2), and what is $x=$ to in (3)? You say "If one of these parts of the equation is $0$..", try to be specific. Rewrite out what equals $0$. – Daniel Buck Sep 13 '16 at 22:17