Matematické Fórum

stuart clark · 22. 01. 2012 17:58

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

where $0<x<2\pi$

stuart clark

i have Triyed like this way

$\sec x+\csc x=c$

$\frac{1}{\cos x}+\frac{1}{\sin x} = c\Leftrightarrow \frac{\sin x+\cos x}{\sin x.\cos x} = c$

$2\left(\sin x+\cos x\right) = c.\sin 2x\Leftrightarrow 4(1+\sin 2x) = c^2\sin ^2 (2x)$

Now Let $\sin (2x) = t$ , Where $-1 \leq t\leq 1$

$c^2t^2-4t-4=0$

So $t=\frac{4\pm 4\sqrt{1+c^2}}{2c^2}=\frac{2\pm 2\sqrt{1+c^2}}{c^2}$

So $\sin (2x) = \frac{2\pm 2\sqrt{1+c^2}}{c^2}$

Now Here $0<x<2\pi$

Now If we divide Given Interval into $4$ Parts, because $\sin x$ and $\cos x$ are different for $4$ parts

Like $x\in \left(0,\frac{\pi}{2}\right)\;\;, x\in \left(\frac{\pi}{2},\pi\right)\;\;, x\in \left(\pi,\frac{3\pi}{2}\right)\;\;,x\in \left(\frac{3\pi}{2},2\pi\right)$

Now for $x\in \left(0,\frac{\pi}{2}\right)$

$\sin (2x)>0$ . So $\sin (2x)<1$

So We take $\sin (2x) = \frac{2\pm 2\sqrt{1+c^2}}{c^2}<1$

From Here We Get $c^2>8$

Now is it Right OR Not help Required

Thanks