I was solving a simple trigonometric equation and found two ways to solve it: one is correct and pretty straightforward, and the other one is a bit more complicated and is really close to being correct, but has extra solutions that came out of nowhere (or at least I think so).

We started doing trigonometry about 3 months ago, so don't be too mad at me if I do some silly mistakes :)

Here's the equation (we need to find $x$): $\sqrt2\sin(x) - \sqrt2\cos(x) = \sqrt3$

Here's the incorrect solution:

$\sqrt2\sin(x) - \sqrt2\cos(x) = \sqrt3$

$\sin(x) - \cos(x) = \frac{\sqrt3}{\sqrt2}$

$1 - 2\sin(x)\cos(x) = 1.5$

$2\sin(x)\cos(x) = -0.5$

$\sin(2x) = -0.5$

$2x = (-1)^{n+1}\arcsin(\frac{1}{2}) + \pi n$

$x = (-1)^{n+1}\frac{\pi}{12}+\frac{\pi n}{2}$, where $n \in\mathbb {Z}$

I don't know why, but after I raise both sides of $\sin(x) - \cos(x) = \frac{\sqrt3}{\sqrt2}$ to the power of 2 (lines 2-3), extra solutions for $x$ appeared.

Here's what I mean:

This is the graph before I raised both sides of my equation to the power of 2

And this is after

As you can see, after I've risen both sides of my equation to the power of 2 extra solutions for $x$ have appeared. And I don't know why that happened. Can somebody please explain this?

Thanks in advance.

P.S. Here's the correct solution:

$\sqrt2\sin(x) - \sqrt2\cos(x) = \sqrt3$

$\frac{\sqrt2}{2}\sin(x) - \frac{\sqrt2}{2}\cos(x) = \frac{\sqrt3}{2}$

$\cos(\frac{\pi}{4} + x) = -\frac{\sqrt3}{2}$

$x = \pm\frac{5\pi}{6} - \frac{\pi}{4} + 2\pi n$, where $n \in\mathbb {Z}$

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    $\begingroup$ Look at it like this $a^2=b^2$ has two solutions $a=\pm b$ now if you have $a=b$ then squaring you get $a^2=b^2$ which has solutions $a=\pm b$ $\endgroup$ – kingW3 Mar 5 '17 at 12:57

When you have $f(x)=g(x)$ and you raise them to the power of $2$, then note that the solutions of $f(x)=-g(x)$ are also solutions of the new equation $[f(x)]^2=[g(x)]^2$. That's why new solutions appear.

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  • $\begingroup$ Ah, I see. So squaring $2\sin(x)\cos(x) = -0.5$ was illegal... $\endgroup$ – Michael Naifield Mar 5 '17 at 13:00
  • $\begingroup$ It's not illegal. You can solve $f(x)=g(x)$ by solving $[f(x)]^2 = [g(x)]^2$, and then, check which solutions correspond to $f(x)=g(x)$ and which to $f(x)=-g(x)$. In other words, it's a correct process which has the drawback of giving extra solutions which do not correspond to your original equation. $\endgroup$ – A. Salguero-Alarcón Mar 5 '17 at 13:07
  • $\begingroup$ Thank you, that made it much more clear. $\endgroup$ – Michael Naifield Mar 5 '17 at 13:14

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