# How to solve this trigonometric equation $\sin (x)-\cos (x)=0$? [closed]

$\sin (x)-\cos (x)=0$. Should I square them both or what should I do?

• several ways to proceed: transform into $\tan(x) = 1$ or $\sin^2(x) = \cos^2(x) = 1/2$ Apr 18 '17 at 19:43
• $sin(x) = cos(x)$, square them, then use identity $1=cos^2(x)+sin^2(x)$
– Χpẘ
Apr 18 '17 at 19:44
• See the second answer of this question: math.stackexchange.com/questions/878279/… Apr 18 '17 at 19:44
• Can you write what you get after squaring and how far you can get from there? That would be one of the possibilities how to add context. (You may have notice that your question received several close votes.) Apr 19 '17 at 3:02

Recall that $\sin(x) = \cos(\frac \pi 2 - x)$. Therefore your equation is: $$\cos(x) = \cos(\frac \pi 2 - x)$$ And from here we get: $x = \frac \pi 2 - x + 2\pi k \Rightarrow 2x = \frac \pi 2 + 2 \pi k \Rightarrow x = \frac \pi 4 + \pi k$ (Of course $k \in \mathbb{Z})$

• How are you sure you don't miss solutions. $\cos (\frac{\pi}{3})=\cos (-\frac{\pi}{3})$ but $\frac{\pi}{3} \neq -\frac{\pi}{3}+2\pi k$. So what I'm trying to say is that in general $\cos u=\cos v$ does not mean $u=v+2\pi k$. Apr 19 '17 at 3:12
• You are right. In general, $\cos u = \cos v \Rightarrow u = v + 2 \pi k$ or $u = -v + 2 \pi k$, but in our case the other equality with the minus sign gives a false statement ($0 = - \frac \pi 2 + 2 \pi k$) Apr 19 '17 at 7:47

Plot the circle $t \mapsto (\cos t, \sin t)$ and see where it intersects the line $y-x = 0$.

Squaring could potentially introduce extraneous solutions, so I recommend avoiding that if possible.

$\sin x - \cos x = 0$ means $\sin x = \cos x$. Now we can note that $\cos x \ne 0$ if $\sin x = \cos x$, because cosine and sine are never zero at the same time (i.e., for the same values of $x$). So it's safe to divide both sides by $\cos x$, and we get $$\tan x = 1.$$

Solving this boils down to knowing the unit circle. Knowing the periodic property of tangent also helps.

If $\sin{x} - \cos{x} = 0$, then $\sin{x} = \cos{x}$. Sine and cosine are both $\pm \frac{\sqrt{2}}{2}$ for some multiple of $\pi/4$ for the angle, and can only intersect there. They are both positive in the first quadrant, and both negative in the third quadrant, so the general solution is $x = \pi/4 + {\pi}n, \ \forall n \in \mathbb{Z}$.

Or, you can divide by $\cos{x}$ on each side, to get $\tan{x} = 1$. Tangent is positive in the first and third quadrants.

If you love squaring things, we have $\sin^2{x} = \cos^2{x}$ and $\sin^2{x} + \cos^2{x} = 1$, so you can now substitute your least favorite function of the two to get: $1 - \cos^2{x} = \cos^2{x} \implies \cos^2{x} = \frac{1}{2} \implies \cos{x} = \pm \frac{\sqrt{2}}{2}$. Not all these solutions work, but if sine was your favorite function, then you would have gotten $\sin{x} = \pm \frac{\sqrt{2}}{2}$ instead. Now we have $\sin{x} = \cos{x} = \pm \frac{\sqrt{2}}{2}$.

As you can see, there are many ways to solve this! There are even four other answers above this one! So when in doubt, really just try something, because when you try something, you are one step closer to solving it, even if you have no idea what you are doing.

$$\sin { x } =\cos { x } \Rightarrow \tan { x } =1\Rightarrow x=\frac { \pi }{ 4 } +n\pi .n\in Z\quad$$