<s>[INST] Is $ \sin 2x \ge 0 $ same as $\sin x\cos x\ge 0 $? Or differently phrased do $$ \sin 2x \ge 0$$  $$\sin x\cos x\ge 0 $$   have the same solution.  
$\sin 2x$ is positive from  $ 0\le x \le \pi/2$ , while $\sin x\cos x$ are positive in the first and third quadrant. Why is this, when $\sin 2x=2\sin x\cos x$, am i making a mistake.  
Does this mean i cannot use double angle identity under the square root?
Edit: as vadim123 said the question was wrong, so i changed it
 [/INST] Note if $M > 0$ then $M*a*b > 0$ if $a > 0; b>0$ or $a < 0; b< 0$.  And $M*a*b < 0$ if $a < 0; b > 0$ or $a > 0; b < 0$.  And $M*a*b = 0$ if $a = 0$ or $b =0$.  It does not matter whether $M =1$ or $M = 2$ or $M = \frac {37.95}{\pi}$.
So we would expect $\cos x \sin x \ge 0$ for the exact same values of $x$ as we would expect $\sin 2x = 2\cos x \sin x\ge 0$ or for the exact same values we would expect $\frac {37.95}{\pi}\cos x \sin x \ge 0$.
And it works out to be exactly the case.  $\cos x \sin x \ge 0$ if $x$ is in the first or third quadrant (as those are the two quadrants where $\sin x$ and $\cos x$ are both the same sign.)  And $\sin M \ge 0$ if $M$ is in the first or second quadrant.  And  $M = 2x$ is in the first quadrant if $x$ is in the first half of the first or third quadrant.  ANd $M$ is in the second quadrant if $x$ is in the second half of the first or third quadrant.
Or in other words:
If $0 \le x \le \frac \pi 2$ then $\sin x \ge 0$ and $\cos x \ge 0$ so $\sin x\cos x \ge 0$.  Also $0 \le 2x \le \pi$ so $\sin 2x \ge 0$.
If $\frac \pi 2 \le x \le \pi$ then $\sin x \ge 0$ and $\cos x \le 0$ so $\sin x\cos x \le 0$.  Also $\pi \le 2x \le 2\pi$ so $\sin 2x \le 0$.
If $\pi \le x \le \frac {3\pi} 2$ then $\sin x \le 0$ and $\cos x \le 0$ so $\sin x\cos x \ge 0$.  Also $2\pi \le 2x \le 3\pi$ so $\sin 2x \ge 0$.
If $\frac {3\pi} 2 \le x \le 2\pi$ then $\sin x \le 0$ and $\cos x \ge 0$ so $\sin x\cos x \le 0$.  Also $3\pi \le 2x \le 4\pi$ so $\sin 2x \le 0$.
As well $\sin 2x = 2\cos x\sin x$  and $\cos x\sin x$ and $M*\cos x \sin x; M > 0$ should all be $> 0$ if $\cos x, \sin x$ are the same sign, and should all be $< 0$ of $\cos x, \sin y$ are opposite signs, and should $= 0$ if one or the other of $\cos x, \sin x$ equals $0$.  So all turns out to work exactly as it should.
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