Find x in [0,2π] to Solve Inequality

[anemone]

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]

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 $\chi$ $\sigma$

 

[kaliprasad]

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/4edited the solution as there was a mistake and it is corrected

 

[I like Serena]

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.

 

[CellCoree]

To solve this inequality, we first need to simplify the expression on the right side of the inequality. By squaring both sides and expanding, we get:
1+2\sin(2x) - 2\sqrt{1-\sin^2(2x)} = 1+2\sin(2x) - 2\cos(2x)
Using the identity \sin^2(x) + \cos^2(x) = 1, we can further simplify this to:
1+2\sin(2x) - 2\cos(2x) = 1+2\sin(2x) - 2\sqrt{1-\cos^2(2x)} = 1+2\sin(2x) - 2|\cos(2x)|
Now, we can rewrite the entire inequality as:
2\cos(x) \le 1+2\sin(2x) - 2|\cos(2x)| \le \sqrt{2}

Next, we can break this into two separate inequalities:
1) 2\cos(x) \le 1+2\sin(2x) - 2|\cos(2x)|
2) 1+2\sin(2x) - 2|\cos(2x)| \le \sqrt{2}

For the first inequality, we can use the identity \cos(2x) = 1-2\sin^2(x) to simplify it to:
2\cos(x) \le 1+4\sin^2(x) - 2|\cos(2x)|
Using the fact that \sin^2(x) \le 1, we can rewrite this as:
2\cos(x) \le 1+4 - 2|\cos(2x)| = 5-2|\cos(2x)|

For the second inequality, we can use the identity \cos(2x) = 1-2\sin^2(x) to rewrite it as:
1+4\sin^2(x) - 2|\cos(2x)| \le \sqrt{2}
Simplifying, we get:
4\sin^2(x) - 2|\cos(2x)| \le \sqrt{2}-1

Now, we can solve each of these inequalities separately. For the first inequality, we can see