Equivalence of solutions of a disequation

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Ozma

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Hi, I must solve [MATH]\log_{2}^{2} \cos x+\log_{2} \cos x \geq 0[/MATH].
My work is this: it must be [MATH]\cos x >0 \Leftrightarrow -\frac{\pi}{2} +2k\pi < x < \frac{\pi}{2}+2k\pi[/MATH]. The disequation is equivalent to
[MATH]\log_{2} \cos x (\log_{2} \cos x +1) \geq 0[/MATH]So I have to study
[MATH]\log_{2} \cos x \geq 0 \Leftrightarrow \cos x \geq 1 \Leftrightarrow x=2k\pi[/MATH]And
[MATH]\log_{2} \cos x \geq 0+1 \geq 0 \Leftrightarrow \cos x \geq \frac{1}{2} \Leftrightarrow -\frac{\pi}{3}+2k\pi \leq x \leq \frac{\pi}{3}+2k\pi[/MATH]So the disequation is positive when [MATH]\left(x \leq -\frac{\pi}{3}+2k\pi \ \text{or} \ x \geq \frac{\pi}{3}+2k\pi\right) \ \text{and} \ x=2k\pi[/MATH]; so considering the existence condition of the logarithms the solutions is [MATH]\left(-\frac{\pi}{2} +2k\pi < x \leq -\frac{\pi}{3}+2k\pi \ \text{or} \ \frac{\pi}{3}+2k\pi \leq x < \frac{\pi}{2}+2k\pi \right) \ \text{and} \ x=2k\pi[/MATH], with [MATH]k\in\mathbb{Z}[/MATH].
However, according to my textbook the solution is [MATH]\left(\frac{\pi}{3}+2k\pi \leq x < \frac{\pi}{2}+2k\pi \ \text{or} \ \frac{3}{2}\pi+2k\pi < x \leq \frac{5}{3}\pi+2k\pi \right) \ \text{and} \ x=2k\pi[/MATH], with [MATH]k\in\mathbb{Z}[/MATH]; am I doing something wrong or are my solution and the textbook's one equivalent somehow? If they are equivalent, why is that so? Is somehow related to the fact that [MATH]k[/MATH] varies in an infinite set so I can obtain one from another letting [MATH]k[/MATH] vary?
Thanks.
 
Ozma said:
Hi, I must solve [MATH]\log_{2}^{2} \cos x+\log_{2} \cos x \geq 0[/MATH].
My work is this: it must be [MATH]\cos x >0 \Leftrightarrow -\frac{\pi}{2} +2k\pi < x < \frac{\pi}{2}+2k\pi[/MATH]. The disequation is equivalent to
[MATH]\log_{2} \cos x (\log_{2} \cos x +1) \geq 0[/MATH]So I have to study
[MATH]\log_{2} \cos x \geq 0 \Leftrightarrow \cos x \geq 1 \Leftrightarrow x=2k\pi[/MATH]And
[MATH]\log_{2} \cos x \geq 0+1 \geq 0 \Leftrightarrow \cos x \geq \frac{1}{2} \Leftrightarrow -\frac{\pi}{3}+2k\pi \leq x \leq \frac{\pi}{3}+2k\pi[/MATH]So the disequation is positive when [MATH]\left(x \leq -\frac{\pi}{3}+2k\pi \ \text{or} \ x \geq \frac{\pi}{3}+2k\pi\right) \ \text{and} \ x=2k\pi[/MATH]; so considering the existence condition of the logarithms the solutions is [MATH]\left(-\frac{\pi}{2} +2k\pi < x \leq -\frac{\pi}{3}+2k\pi \ \text{or} \ \frac{\pi}{3}+2k\pi \leq x < \frac{\pi}{2}+2k\pi \right) \ \text{and} \ x=2k\pi[/MATH], with [MATH]k\in\mathbb{Z}[/MATH].
However, according to my textbook the solution is [MATH]\left(\frac{\pi}{3}+2k\pi \leq x < \frac{\pi}{2}+2k\pi \ \text{or} \ \frac{3}{2}\pi+2k\pi < x \leq \frac{5}{3}\pi+2k\pi \right) \ \text{and} \ x=2k\pi[/MATH], with [MATH]k\in\mathbb{Z}[/MATH]; am I doing something wrong or are my solution and the textbook's one equivalent somehow? If they are equivalent, why is that so? Is somehow related to the fact that [MATH]k[/MATH] varies in an infinite set so I can obtain one from another letting [MATH]k[/MATH] vary?
Thanks.
Click to expand...
Did you try graphing the two solutions to see how they compare?

What you'll see is that your

[MATH]-\frac{\pi}{2} +2k\pi < x \leq -\frac{\pi}{3}+2k\pi[/MATH]​

is equivalent, by replacing [MATH]k[/MATH] with [MATH](k+1)[/MATH], to

[MATH]-\frac{\pi}{2} +2(k+1)\pi < x \leq -\frac{\pi}{3}+2(k+1)\pi[/MATH],​

which is

[MATH]2\pi-\frac{\pi}{2} +2k\pi < x \leq 2\pi-\frac{\pi}{3}+2k\pi[/MATH],​

and therefore

[MATH]\frac{3\pi}{2} +2k\pi < x \leq \frac{5\pi}{3}+2k\pi[/MATH],​

which is their version. The rest is all identical.
 
@Dr.Peterson: First of all thanks for your clear answer. Yes, I did a graph and I thought that I can obtain one from another shifting [MATH]k[/MATH] as you did but I was not sure, thank you for doing the calculations explicitly!
In general if I have a disequation with periodic function it is enough to find the interval of the solutions in one period and then extend the solution adding the periodicity?
 
Ozma said:
@Dr.Peterson: First of all thanks for your clear answer. Yes, I did a graph and I thought that I can obtain one from another shifting [MATH]k[/MATH] as you did but I was not sure, thank you for doing the calculations explicitly!
In general if I have a disequation with periodic function it is enough to find the interval of the solutions in one period and then extend the solution adding the periodicity?
Click to expand...
Yes. It only becomes difficult when the arguments of trig functions are not merely x, and especially when they are not the same.

Parts of your work, however, seem wrong.

I made a substitution of [MATH]u = \log_2(\cos(x))[/MATH], so that [MATH]u(u+1)\ge 0[/MATH], and [MATH]u\le -1[/MATH] or [MATH]u\ge 0[/MATH]. Therefore [MATH]\log_2(\cos(x))\le -1[/MATH] or [MATH]\log_2(\cos(x))\ge 0[/MATH], so that [MATH]0<\cos(x)\le \frac{1}{2}[/MATH] or [MATH]\cos(x)\ge 1[/MATH]. Then I sketched a graph of the cosine, marking where it is 1 or between 0 and 1/2.

But you say [MATH]\cos x \geq \frac{1}{2}[/MATH], which is wrong, and then [MATH]\left(x \leq -\frac{\pi}{3}+2k\pi \ \text{or} \ x \geq \frac{\pi}{3}+2k\pi\right)[/MATH], which is very wrong, as it is not restricted to a cycle. So it looks as if your correct answer may be an accident!
 
Dr.Peterson said:
Yes. It only becomes difficult when the arguments of trig functions are not merely x, and especially when they are not the same.
Click to expand...

@Dr.Peterson: In that case I have to find the common period of all the trigonometric functions involved and then add that period in the interval of the solutions?

Dr.Peterson said:
Parts of your work, however, seem wrong.

I made a substitution of [MATH]u = \log_2(\cos(x))[/MATH], so that [MATH]u(u+1)\ge 0[/MATH], and [MATH]u\le -1[/MATH] or [MATH]u\ge 0[/MATH]. Therefore [MATH]\log_2(\cos(x))\le -1[/MATH] or [MATH]\log_2(\cos(x))\ge 0[/MATH], so that [MATH]0<\cos(x)\le \frac{1}{2}[/MATH] or [MATH]\cos(x)\ge 1[/MATH]. Then I sketched a graph of the cosine, marking where it is 1 or between 0 and 1/2.

But you say [MATH]\cos x \geq \frac{1}{2}[/MATH], which is wrong, and then [MATH]\left(x \leq -\frac{\pi}{3}+2k\pi \ \text{or} \ x \geq \frac{\pi}{3}+2k\pi\right)[/MATH], which is very wrong, as it is not restricted to a cycle. So it looks as if your correct answer may be an accident!
Click to expand...
Oh yes, when I've written that I was referring to the study of the sign of the multiplication [MATH]\log_{2} \cos x (\log_{2} \cos x +1)[/MATH], my teachers taught me that I must study when the terms of the product are positive alone and then do a graph to determine when the product is positive or negative all together (that's why I've written both [MATH]\geq 0[/MATH], sorry if it was confused).
So from the graph the product is positive, as you've said, when [MATH]\cos x \geq 1[/MATH] or [MATH]\cos x \leq \frac{1}{2}[/MATH].

Thank you.
 
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