Question

How many solutions does the equation $\sec x-1=(\sqrt{2}-1)\tan x$ have in the interval $(0,6\pi ]$?

• A. $6$
• B. $5$
• C. $10$
• D. $9$

Answer 1

Answer: (A)

$6$

$\sec x-1=(\sqrt{2}-1)\tan x$

$\frac{1}{\cos x}-1=(\sqrt{2}-1)\frac{\sin x}{\cos x}$

$1-\cos x=(\sqrt{2}-1)\sin x$

Applying $1-\cos x=2{\sin }^2\left(\frac{x}{2}\right)$ and $\sin x=2\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)$ in above equation$:$

$\Rightarrow 2{\sin }^2\left(\frac{x}{2}\right)=(\sqrt{2}-1)2\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)$

$\Rightarrow {\sin }^2\left(\frac{x}{2}\right)-(\sqrt{2}-1)\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)=0$

$\Rightarrow \sin \left(\frac{x}{2}\right)\left(\sin \right(\frac{x}{2})-(\sqrt{2}-1\left)\cos \right(\frac{x}{2}\left)\right)=0$

$\Rightarrow \sin \left(\frac{x}{2}\right)=0$ or $\tan \left(x/2\right)=(\sqrt{2}-1)$

$\tan \left(x/2\right)=(\sqrt{2}-1)\Rightarrow \tan \left(x/2\right)=\tan (\pi /8+\pi n)$

$\Rightarrow x/2=\pi /8+\pi n$

$x=\pi /4+2\pi n,n$ is an integer

Given $xϵ(0,6\pi ]$

$\therefore x=\frac{\pi }{4},\frac{\pi }{4}+2\pi ,\frac{\pi }{4}+4\pi$

$\sin \left(\frac{x}{2}\right)=0\Rightarrow x/2=n\pi$

$x=2n\pi ,n$ is an integer

$\therefore x=2\pi ,4\pi ,6\pi$

That is $6$ solutions in total.

Hence, $\left(A\right)$