# Find all the actual $x$ values that satisfy the following inequality

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:

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

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

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

and then put togheter the intervals case by case.

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