in the interval of $0< \theta <360$ solve for $\theta$ , the equation $(1)$, I would just say $\tan \theta =-1$ and solve that.


but what if i said $\sin θ +\cos θ =0$ then $\cos θ (\tan θ +1)=0$. I'd have solutions for $\cos θ =0$ as well as $\tan θ =-1$ , why is doing this method wrong since arithmetically going through the workings out it seems fine , but the solutions for $\cosθ=0$ are wrong?

I'm trying to get a verdict on if the 2nd method is still right or wrong ? i have a feeling its wrong since the solutions dont work, but the process of me getting to $\cos(θ)⋅[\tanθ+1]=0$ seemed sound that I wouldnt realise it was wrong ?

How do I stop myself from carrying out this sort of working out ,is there a way for me to intuitively prove its wrong even before the equation ends up in the form of $\cos(θ)⋅[\tanθ+1]=0$ because if i carried out a similar working out for this question " Solve for theta ,$\sinθ=3\sinθ\cosθ$" and solving it similarly like so I'd end up with $\sinθ(1-3\cosθ)=0$ ,and it would give the right solutions.

  • 1
    $\begingroup$ In the 1st method, you divided both sides by $\cos\theta$, which is illegitimate if $\cos\theta=0$. $\endgroup$ Jan 20 at 5:21
  • $\begingroup$ In the 2nd method, you converted $\sin\theta = \cos\theta\tan\theta$, which is undefined when $\cos\theta = 0$. $\endgroup$
    – peterwhy
    Jan 20 at 5:25
  • 1
    $\begingroup$ Either method, when you choose to divide by $\cos\theta$, consider the two cases: either 1) $\cos\theta = 0$ then no solutions, or 2) $\cos\theta \ne 0$ then you can safely divide. $\endgroup$
    – peterwhy
    Jan 20 at 5:40
  • $\begingroup$ @peterwhy so you are saying whenever i divide by a trig function consider if that trigfunction=0 would there be solutions to it or not , so in the case of cosθ=0 there wouldnt be solutions either way , so either method i did would be right ?also when you say i chose to divde by cosθ in the 2nd scenario you mean when i pulled out cosθ to make a set of factors cos(θ)⋅[tanθ+1]=0 $\endgroup$
    – j jose
    Jan 20 at 6:33
  • 1
    $\begingroup$ @jjose: There's an important nuance here. The rule that allows us to say $ab=0$ when $a=0$ or $b=0$ pre-supposes that $a$'s zero-ness doesn't cause trouble for $b$ (or vice-versa). Your second strategy is like solving $x-3=0$ by factoring to get $x\left(1-\frac3x\right)=0$, and then claiming $x=0$ is a soln because it makes the first factor vanish; however, $x=0$ makes the second factor undefined, which is problematic. (Moreover, if that strategy worked, then $0$ would be a soln to every eqn!) You're calling attention to the fact that we can't consider each factor in complete isolation. $\endgroup$
    – Blue
    Jan 21 at 11:51

2 Answers 2


In the first method,

$${\sin\theta=-\cos\theta\\\implies \begin{cases} \tan x=-1,& \text{if } \cos \theta\ne 0\\ \sin\theta=0, & \text{if } \cos \theta= 0\rightarrow\text{No solution.}\text{ As, $\cos \theta$ and $\sin \theta$ both cannot $0$ simultaniously. } \end{cases}\\} $$

Hence, your first method is right.

In the second method,

When we take out common factor $\cos\theta$, we are dividing $\sin\theta$ by $\cos\theta$ so we have assumed that $\cos\theta\ne0$, $$\sin\theta+\cos\theta=0{\implies\cos \theta (\tan \theta +1)=0\quad \text{When $\cos\theta\ne0$}\\\implies \tan\theta=-1\quad(\because \cos\theta\ne0)}$$

Edit: We will prove that $\sin\theta+\cos\theta=\cos \theta (\tan \theta +1)$ is invalid when $\cos\theta=0$:

LHS$=\pm1+0=\pm1$ but, RHS$=0\times(\text{undefined}+1)=\text{undefined}$

But, in the case $\sinθ-3\sinθ\cosθ=\sinθ(1−3\cosθ)$ it is valid for $\sin\theta=0$:

LHS$=0-3\times0\times(\pm1)=0$ and, RHS$=0\times(1-3\times(\pm1))=0$

Alternative aaproach,

$$\sin\theta+\cos\theta=0\\ {\implies\sqrt2\sin\left(\theta+\frac{\pi}{4}\right)=0\\ \implies\theta+\frac{\pi}{4}=n\pi \quad\text{Where, } n\in\Bbb Z}$$

  • $\begingroup$ you mentioned how when i factor out cosθ i assume its not equal to 0 , which in many questions will be seen as a mistake as you will miss out on solutions , so how am i supposed to know when to divide by a trig function and if i will still have all the solutions even when i divde through by a trig . Like in this case sinθ=3sinθcosθ goes to sinθ(1−3cosθ)=0 , if i were to divide by sin rather than factoring it out i would have lost solutions as when you divided by sinθ you assumed its not=0 $\endgroup$
    – j jose
    Jan 20 at 20:14
  • $\begingroup$ @jjose No. You diving $\sin\theta$ by $\cos\theta$ to get $\tan \theta$ so, . If something like $\sin θ-3\sin θ\cos θ=0$ then you can factor out $\sin θ(1−3\cos θ)=0$ here you not dividing any thing. I edited my answer and add your query $\endgroup$
    – O M
    Jan 21 at 9:23

As noticed with the first method the equivalence

$$\sin\theta=-\cos\theta \iff \cos \theta (\tan \theta +1)=0$$

is true only when $\cos \theta \neq 0$ and since $\cos \theta=0$ is not a solution to the original equation, indeed

$$\cos \theta =0 \implies \sin\theta=\pm 1$$

therefore we can solve for $$\tan \theta +1=0$$

As another alternative we have

$$\sin\theta=-\cos\theta \iff \cos \left(\frac \pi 2 -\theta\right)=\cos(\pi\pm\theta)$$

then use

$$\cos x = \cos y \iff x=y+2k\pi \;\lor\; x=-y+2k\pi$$

which lead to

  • $\frac \pi 2 -\theta=\pi\pm\theta+2k\pi \implies \theta=-\frac \pi 4+k\pi$
  • $\frac \pi 2 -\theta=-\pi\mp\theta+2k\pi \implies \theta=-\frac 3 4\pi+k\pi$

and the solutions are $x=135°$ and $x=315°$.

Another way by squaring

$$\sin\theta=-\cos\theta \iff \sin\theta+\cos\theta=0 \implies \sin 2\theta =-1$$

which leads to

$$2\theta =\frac 3 2 \pi +2k\pi \implies \theta = \frac 3 4 \pi +k\pi$$

The case $\sin\theta=3\sin\theta\cos\theta$ is slightly different, here we can factor out as follows (this is always true, we have no limitations on $\theta$)

$$\sin\theta=3\sin\theta\cos\theta \iff \sin\theta(1-3\cos\theta)=0$$

and the latter is equivalent to

$$\sin \theta =0 \;\lor \; 1-3\cos\theta=0$$

  • $\begingroup$ @jjose Thanks I’ve fixed it! Hope now it is clear. $\endgroup$
    – user
    Jan 21 at 11:48
  • $\begingroup$ :505767 how does cosθ=0 imply that sinθ=+ or - 1 and therfore mean that tanθ+1? sorry im not great with maths speak are you saying that when you divide cosθ when you have sinθ=-cosθ or when you factors out cosθ from sinθ+cosθ=0 , you wont lose solutions as you would if you were to divide sinθ=3sinθcosθ by sinθ because in this latter case you are canceling sinθ from the equation ,but in the former cases you are not $\endgroup$
    – j jose
    Jan 21 at 11:55
  • $\begingroup$ im trying to see how the case of sinθ=3sinθcosθ⟺sinθ(1−3cosθ)=0 instead of dividing by sinθ from the get go is different to sinθ=-cosθ >>> dividing cosθ from the get go to solve which is allowed here but not allowed in the former case. $\endgroup$
    – j jose
    Jan 21 at 11:57
  • $\begingroup$ There are many ways to manipulate and solve the equation, each one requires to pay attention. $\endgroup$
    – user
    Jan 21 at 12:00
  • $\begingroup$ For $\sin x=-\cos x$ we can note that $\cos x=0$ is not a solution, then we can assume $\cos x\neq 0$ and divide the original equation by $\cos x$ to obtain $\tan x=-1$ which is equivalent to the original problem. $\endgroup$
    – user
    Jan 21 at 12:02

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