0
$\begingroup$

Knowing that $ i ^ 2 = -1 $, find all the actual $ x $ values that satisfy the following inequality: $$Re \left\{\frac{2 \log_2 \sin(x)+1}{i \left(e^{2ix}-2 \cos^2(x)+1 \right)}\right\}>0$$Where $ Re $ {$ Z $} is the real part of the $ Z $ complex number

I did the reduced form:

enter image description here

$\endgroup$
  • $\begingroup$ You are right! Do you also need some advice for the inequality? $\endgroup$ – user Dec 5 '19 at 20:27
  • $\begingroup$ @user Could you tell me what to do until you finish the solution? $\endgroup$ – Benemon Dec 5 '19 at 20:28
  • $\begingroup$ I add something on that too. $\endgroup$ – user Dec 5 '19 at 20:29
2
$\begingroup$

You are right, indeed we have that

$$\frac{1}{e^{2ix}-2 \cos^2(x)+1 }=\frac{1}{\cos(2x)-2 \cos^2(x)+1+i\sin (2x) }=\frac{1}{i\sin (2x) }$$

therefore

$$\frac{2 \log_2 \sin(x)+1}{i \left(e^{2ix}-2 \cos^2(x)+1 \right)}=-\frac{2 \log_2 \sin(x)+1}{\sin(2x)}$$

and we need to solve $\frac{2 \log_2 \sin(x)+1}{\sin(2x)}<0$.

Then let consider separately the numerator and the denominator, notably

$$\sin(2x)>0 \iff 2k\pi <2x<\pi+2k\pi $$

$$\sin(2x)<0 \iff \pi+ 2k\pi <2x<2\pi+2k\pi $$

and

$$2 \log_2 \sin(x)+1> 0 \iff \log_2 \sin(x)>-\frac12=\log_2 \frac1{\sqrt 2} \iff \sin x>\frac{\sqrt 2}2 $$$$\iff \frac \pi 4+ 2k\pi <x<\frac {3\pi} 4+2k\pi$$

$$2 \log_2 \sin(x)+1< 0 \iff \log_2 \sin(x)<-\frac12=\log_2 \frac1{\sqrt 2} \iff 0<\sin x<\frac{\sqrt 2}2$$$$\iff 2k\pi <x<\frac {\pi} 4+2k\pi \cup \frac {3\pi} 4+2k\pi <x<\pi+2k\pi$$

and then put togheter the intervals case by case.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Where did you miss? What is missing to complete? $\endgroup$ – Benemon Dec 5 '19 at 20:28
  • $\begingroup$ @Tortugut Now we need to solve the inequality. Can you proceed with that? $\endgroup$ – user Dec 5 '19 at 20:28
  • $\begingroup$ Ok can you show me the steps $\endgroup$ – Benemon Dec 5 '19 at 20:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.