Trigonometric equation - resolver sec(x)+csc(x)=c

[kinu]

Prove that equation \(\sec x+\csc x = c\) has \(2\) solution when \(c^2<8\) and \(4\) solution when \(c^2>8\)

where \(0<x<2\pi\)

[Rui Carpentier]

Just a sketch of a proof:

Let \(f(x)=\sec(x)+\csc(x)\)

Find \(f'\) and show that: \(f'(x)>0\) for \(x\in (\frac{\pi}{4},\frac{\pi}{2})\cup (\frac{\pi}{2},\pi)\cup (\pi,\frac{5\pi}{4})\) ; \(f'(x)<0\) for \(x\in (0,\frac{\pi}{4})\cup (\frac{5\pi}{4},\frac{3\pi}{2})\cup (\frac{3\pi}{2},2\pi)\) and \(f'(x)=0\) for \(x=\frac{\pi}{4}\vee x=\frac{5\pi}{4}\). Conclude that \(f\) is strictly decreasing on the intervals \((0,\frac{\pi}{4}), (\frac{5\pi}{4},\frac{3\pi}{2})\) and \((\frac{3\pi}{2},2\pi)\) and is strictly increasing on the intervals \((\frac{\pi}{4},\frac{\pi}{2}), (\frac{\pi}{2},\pi)\) and \((\pi,\frac{5\pi}{4})\). Thus it is one-to-one in those intervals.

Calculate \(\lim_{x\rightarrow 0^+}f(x)=+\infty\), \(f(\frac{\pi}{4})=\sqrt{8}\), \(\lim_{x\rightarrow \frac{\pi}{2}^-}f(x)=+\infty\), \(\lim_{x\rightarrow \frac{\pi}{2}^+}f(x)=-\infty\), \(\lim_{x\rightarrow \pi^-}f(x)=+\infty\), \(\lim_{x\rightarrow \pi^+}f(x)=-\infty\), \(f(\frac{\pi}{4})=-\sqrt{8}\), \(\lim_{x\rightarrow \frac{3\pi}{2}^-}f(x)=-\infty\), \(\lim_{x\rightarrow \frac{3\pi}{2}^+}f(x)=+\infty\) and \(\lim_{x\rightarrow 2\pi^-}f(x)=-\infty\).

Then conclude that \(f(x)=c\) has 2 solutions when \(c^2<8\) and 4 solutions when \(c^2>8\).

[kinu]

Thanks Rui