# Inequality challenge

#### anemone

##### MHB POTW Director
Staff member
Find all $$\displaystyle x$$ in the interval $$\displaystyle [0, 2\pi]$$ which satisfies $$\displaystyle 2\cos(x) \le \left|\sqrt{1+\sin (2x)}-\sqrt{1-\sin (2x)} \right|\le \sqrt{2}$$

#### chisigma

##### Well-known member
Find all $$\displaystyle x$$ in the interval $$\displaystyle [0, 2\pi]$$ which satisfy $$\displaystyle 2\cos x \le |\sqrt{1+\sin (2x)}-\sqrt{1-\sin (2x)}|\le \sqrt{2}$$
If correct, the problem is set up as 'non challenge question' because the requirement is reduced to...

$\displaystyle \cos x \le \frac{1}{\sqrt{2}}\ (1)$

... and the (1) is satisfied for $\displaystyle \frac{\pi}{4} \le x \le \frac{7}{4} \pi$...

Kind regards

$\chi$ $\sigma$

##### Well-known member
Find all $$\displaystyle x$$ in the interval $$\displaystyle [0, 2\pi]$$ which satisfies $$\displaystyle 2\cos(x) \le \left|\sqrt{1+\sin (2x)}-\sqrt{1-\sin (2x)} \right|\le \sqrt{2}$$
one solution cos x <= 0

so x is between pi/2 and - 3pi/2

second case cos x > = 0 and sin x >= 0

now sin x < cos x gives
cos x < 1/2 | cos x + sin x - ( cos x - sin x) <= 1/ sqrt(2)

or cos x < sin x <= 1/sqrt(2)

cos x >= 1/sqrt(2) => sin x >= 1/ sqrt(2) so no solution

cos x > 0 and sin x >= cos x gives cos x <= sqrt(2)

siimiliarly ranges sin x < 0 2 ranges | sin x | < cos x and | sin x | > cos x need to be anlaysed

by symetry we get cos <= 1/ sqrt(2)

so solution set cos^1 (1/ 2sqrt(2) to 2pi - cos^1 (1/2sqrt(2))
or between pi/4 and 7pi/4

edited the solution as there was a mistake and it is corrected

Last edited:

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Find all $$\displaystyle x$$ in the interval $$\displaystyle [0, 2\pi]$$ which satisfies $$\displaystyle 2\cos(x) \le \left|\sqrt{1+\sin (2x)}-\sqrt{1-\sin (2x)} \right|\le \sqrt{2}$$
Squaring the expression in the middle gives:
\begin{aligned}\left|\sqrt{1+\sin (2x)}-\sqrt{1-\sin (2x)} \right|^2
&= (1+\sin (2x))+(1-\sin (2x)) - 2 \sqrt{(1+\sin (2x))(1-\sin (2x))} \\
&= 2 - 2 \sqrt{1-\sin^2(2x)} \\
&= 2 - 2 \sqrt{\cos^2(2x)} \\
&= 2 - 2 |\cos(2x)| \\
&= 2 - 2 |2\cos^2 x - 1|
\end{aligned}

If $\cos(2x)> 0$, this reduces to $4 - 4 \cos^2 x$.
If $\cos(2x)\le 0$, this reduces to $4 \cos^2 x$.

The latter case is a 1-1 match for the squared left hand side.
Checking out the cases where $\cos x > 0$ respectively $\cos x \le 0$ yields that the first inequality is always true.

chisigma already gave the solution when looking at the remaining inequalities.