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$$\cos x - \sin x = -1$$

There are 2 methods to solve the equation:

  1. Dividing by $\sqrt{2}$ to get $\cos (\frac{\pi}{4} + x) = \cos(\frac{3 \pi}{4} \rightarrow x=(2n(\pi)+ \frac{\pi}{2} ~\text{or} ~((2n-1)\pi)$

  2. Squaring to get $\sin 2x = 0 \rightarrow x=\frac{n(\pi)}{2}$

Are both the solutions true and why does such a situation arise?

Thanks in advance

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When you square an equation you often introduce extraneous roots. The simplest example is $x=1$. If you square it, you get $x^2=1$, which is also satisfied by $x=-1$. Squaring is non-reversible, so you need to check solutions in the original equation if you use it.

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But then why dont I get the original solution along with the extraneous one? Like in the example u gave u also get 1 as solution along with -1 – user60576 Feb 4 '13 at 17:05
@user60576: It does include your other solution. For example, your first solution $n=1$ gives $\frac 52 \pi$ or $\pi$. If you put $5$ for $n$ in the second you get the first of these, if you put $2$ you get the second. – Ross Millikan Feb 4 '13 at 17:12
So is it that some sols of n(pi)/2 satisfy??? yeah !!!! Perfect..... Thanks – user60576 Feb 4 '13 at 17:14
@user60576: Yes, squaring will keep all the solutions, it will just add others. – Ross Millikan Feb 4 '13 at 17:24

No. They are not the same.

In a), you get the correct solution.

In b), while squaring, you are considering the case $\cos x-\sin x=1$ as well, which are will not satisfy your given equation.

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I raise the same doubt as in comment on above ans that why dont i get the true sol. with extraneous one? – user60576 Feb 4 '13 at 17:12
You got the answer, as well from @Ross Millikan :) – Tapu Feb 4 '13 at 17:13

Let's be absolutely clear and transparent: We can solve the equation two ways: $$\sqrt{2}\left(\sin(\pi/4)\cos(x)-\cos(\pi/4)\sin(x)\right)=-1\Longrightarrow \sin(\pi/4-x)=-(1/\sqrt{2}).$$ Simplifying this yields $$\pi/4-x=-\pi/4+2n\pi\Longrightarrow x=\pi/2+2n\pi$$ or$$\pi/4-x=-3\pi/4+2n\pi\Longrightarrow x=(2n-1)\pi.$$

Alternatively, we can square the equation, and we get $$\cos^2(x)-2\sin(x)\cos(x)+\sin^2(x)=1-\sin(2x)=1\Longrightarrow \sin(2x)=0.$$ Again, simplifying, we have $$2x=n\pi\Longrightarrow x=\frac{n\pi}{2}.$$Since we've squared a term here, we have to go back and plug in our terms, but notice that all of the solutions for this first method are included here (precisely for $n=4k+1$), we just happen to have extraneous solutions gained from squaring the equation. Now, we want $$\cos(n\pi/2)-\sin(n\pi/2)=-1.$$ From here, it is easy to see this is precisely when $n\equiv 1\pmod{4}$ since at $n\equiv 3\pmod{4}$ we have the equation equal to $1$, not $(-1)$, or when or $n\equiv 2\pmod{4}$ and not $n\equiv 0\pmod{4}$ for the same reason as before.

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