We have $\sqrt{1+2\sin{2x}}= \cos{x}-\sin{x}$
Fistly my conception was squaring both sides but I figured out that this method is wrong, so my question is how should look solution this example ?
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1$\begingroup$ Squaring both sides is good. We do have to examine the roots we get to see whether they really satisfy the original equation. $\endgroup$– André NicolasCommented Feb 28, 2014 at 19:40
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$\begingroup$ but how to obtain the same result as wolfram $\endgroup$– GregorCommented Feb 28, 2014 at 19:49
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$\begingroup$ Programs and people solve problems in different ways, it is not a good idea to try to emulate Alpha. For one thing, it "likes" to travel through the complex numbers. $\endgroup$– André NicolasCommented Feb 28, 2014 at 19:52
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$\begingroup$ What's wrong with complex numbers? $\endgroup$– NishantCommented Aug 6, 2014 at 20:27
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2 Answers
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Squaring both sides gives
$$1+2\sin2x=\cos^2x-2\cos x\sin x+\sin^2x=1-\sin2x$$
or
$$\sin2x=0$$
This suggest $x=k\pi/2$ for $k\in\mathbb{Z}$, but care must be taken to eliminate the ones for which $\cos x-\sin x=-1$ instead of $+1$. That leaves $2k\pi$ and $(4k-1)\pi/2$ for $k\in\mathbb{Z}$.
answered Feb 28, 2014 at 19:53
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$\begingroup$ why just for $\cos{x}-\sin{x}=-1$ shouldn't be $\cos{x}-\sin{x} \ge 0$ ?? $\endgroup$– GregorCommented Feb 28, 2014 at 19:58
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$\begingroup$ @Gregor, when $x=k\pi/2$, $\cos x-\sin x=\pm1$. $\endgroup$ Commented Feb 28, 2014 at 20:02
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Well why not square it? $$\sqrt{1+2\sin2x}=\cos x-\sin x\\1+2\sin 2x=\cos^2x-2\cos x\sin x+\sin^2x\\1+4\sin x \cos x=1-2\cos x \sin x\\6\sin x\cos x=0 \implies x=\frac{k\pi}{2}$$ NOTE:I used identities $$\sin 2x=2\cos x\sin x\\\sin^2x+\cos^2x=1$$
