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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!

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

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  • $\begingroup$ So $m$ can have only the following values: $(0;{\sqrt3 \over 2})$! $\endgroup$ – I.Gandakov Apr 24 '14 at 11:17
  • $\begingroup$ More or less right idea, but you need to look at the details a bit more carefully. $\endgroup$ – David Apr 24 '14 at 11:20
  • $\begingroup$ I got that $|m|\geq\sqrt{3}/2$... $\endgroup$ – MPW Apr 24 '14 at 11:22
  • $\begingroup$ @JohnG. The bound $\frac {\sqrt 3}{2}$ is correct but the other is not, I am afraid. Cheers. $\endgroup$ – Claude Leibovici Apr 24 '14 at 11:30
  • $\begingroup$ @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:) $\endgroup$ – I.Gandakov Apr 24 '14 at 11:30

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