Question

Determine all the values of $x$ in the interval $xϵ[0,2\pi ]$ which satisfy the inequality $2\cos x\le |\sqrt{1+\sin 2x}-\sqrt{1-\sin 2x}|\le 2$

Answer 1

$2cos\left(x\right)\le |\sqrt{1+sin2x}-\sqrt{1-sin2x}|\le 2$
$2cos\left(x\right)\le |\sqrt{si{n}^2\left(x\right)+co{s}^2\left(x\right)+2sin\left(x\right).cos\left(x\right)}-\sqrt{si{n}^2\left(x\right)+co{s}^2\left(x\right)-2sin\left(x\right).cos\left(x\right)}|\le 2$
$2cos\left(x\right)\le |sin\left(x\right)+cos\left(x\right)-\left(cos\right(x)-sin(x\left)\right)|\le 2$
$2cos\left(x\right)\le |2sin\left(x\right)|\le 2$
$cos\left(x\right)\le |sin\left(x\right)|\le 1$
Now $|sin\left(x\right)|\le 1$ is true for all $xϵ[0,2\pi ]$.
Hence
$|sin\left(x\right)|\ge cos\left(x\right)$.
Or
$sin\left(x\right)\ge cos\left(x\right)$ and $sin\left(x\right)\le -cos\left(x\right)$
Now
$sin\left(x\right)-cos\left(x\right)\ge 0$
$sin(x-\frac{\pi }{4})\ge 0$
$0\le x-\frac{\pi }{4}\le \pi$
$\frac{\pi }{4}\le x\le \frac{5\pi }{4}$ ...(i)
And
$sin\left(x\right)\le -cos\left(x\right)$
$sin\left(x\right)+cos\left(x\right)\le 0$
$sin(x+\frac{\pi }{4})\le 0$
$xϵ[0,\frac{\pi }{4}]\cup [\frac{5\pi }{4},2\pi ]$...(ii)