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 ?
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 ?
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 $k\in\mathbb{Z}$.
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 $$\sin 2x=2\cos x\sin x\\\sin^2x+\cos^2x=1$$