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For which values of real parameter "m" the equation:$$\sqrt3*|\tan x+\cot x|=4m$$ has real solutions? My only thought is that $m\gt 0$ because the right part of the equation is an absolute value which is always positive. That's the only thing I can say. I hope you'll add some more ideas, and help me solve this exercise. Thank you!

Mathias711
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wonderingdev
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1 Answers1

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Hint: $$\tan x+\cot x=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x} =\frac{\sin^2x+\cos^2x}{\cos x\sin x}=\frac{2}{\sin2x}\ .$$

David
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  • So $m$ can have only the following values: $(0;{\sqrt3 \over 2})$! – wonderingdev Apr 24 '14 at 11:17
  • More or less right idea, but you need to look at the details a bit more carefully. – David Apr 24 '14 at 11:20
  • I got that $|m|\geq\sqrt{3}/2$... – MPW Apr 24 '14 at 11:22
  • @JohnG. The bound $\frac {\sqrt 3}{2}$ is correct but the other is not, I am afraid. Cheers. – Claude Leibovici Apr 24 '14 at 11:30
  • @David I noticed that $sin2x$ gives values from -1 to 1 , thus the values of absolute function will be from 0 to 2. This means that the left part of the equation gives values from 0 to $2\sqrt3$. But not to forget, sin2x cannot be $0$ so, $x\neq {\pi\over 4}$. So M can have only values ($0;{\sqrt3\over2}$){$\pi\over 2$}.Is this right? Please comment if it is:) – wonderingdev Apr 24 '14 at 11:30
  • @JohnG., I'm not sure but it looks to me as if you are multiplying when you should be dividing. Or is it possible that there was a typo in your question? Please check carefully and let us know. – David Apr 24 '14 at 11:34
  • Yes, but I am multiplying when I tend to find what values can the left part of the equation get(I multiply by $\sqrt3$, and after that, I divide by 4 the margins of the interval I've got, to find out which are the values of $m$). Isn't that right? – wonderingdev Apr 24 '14 at 11:38
  • What values can $|\sin2x|$ have? What do you get if you divide $\sqrt3/2$ by these values? – David Apr 24 '14 at 11:39
  • |sin2x| can have values only from 0 to 1. |2/sin2x| can have values from 0 to 2 . and finally $\sqrt3|{2\over sin2x}|$ can have values from 0 to $2\sqrt 3$. Sorry for keeping you so much with my exercise. – wonderingdev Apr 24 '14 at 11:42
  • No, $|2\sin2x|$ has values from $0$ to $2$. But $|2/!\sin2x|$ is different. – David Apr 24 '14 at 11:45