Question

Solve the inequality :-

$\sqrt{x - {x}^{2} } \times ( \sin(x) - \frac{1}{2}) \geqslant 0$

Answer 1

Final Answer :
$x= 0 \: and \: \\ \frac{\pi}{6} \leqslant x \leqslant 1$
Steps and Understanding :

1) Domain of LHS function :
$x - {x}^{2} \geqslant 0 \\ = > {x}^{2} - x \leqslant 0 \\ = > x(1 - x) \leqslant 0 \\ = > x \leqslant 1 \: and \: x \geqslant 0$

So, we will restriction ourselves in this domain.

2) Critical points :
$x - {x}^{2} = 0 \\ = > x = 0 \: or \: 1$
$\sin(x) = \frac{1}{2} \\ = > x = \frac{\pi}{6}$

So, now we make number line and give signs.

(3) Since root is always non negative, so forget about it. Our focus is in trigonometric function .
Between π/6 and 1 , we get
sin(x) - 1/2 as positive so we use positive sign between π/6 and 1.
and negative in between 0 and π/6 which is not required in given function.
See, Number line in pic.

(4) Critical points will satisfy as there is equality sign.

We get our solution as :
$\frac{\pi}{6} \leqslant x \leqslant 1 \: or \: x \: = 0$