Problem source

Sketch the graphs of $y=\sec x$ and $y=\ln(2\sec x)$ for $0\leqslant x\leqslant\frac{1}{2}\pi$.
Show graphically that the equation 
\[
kx=\ln(2\sec x)
\]
has no solution with $0\leqslant x<\frac{1}{2}\pi$ if $k$ is a small positive number but two solutions if $k$ is large. 
Explain why there is a number $k_{0}$ such that 
\[
k_{0}x=\ln(2\sec x)
\]
has exactly one solution with $0\leqslant x<\frac{1}{2}\pi$. Let
$x_{0}$ be this solution, so that $0\leqslant x_{0}<\frac{1}{2}\pi$
and $k_{0}x_{0}=\ln(2\sec x_0)$. Show that 
\[
x_{0}=\cot x_{0}\ln(2\sec x_{0}).
\]
Use any appropriate method to find $x_{0}$ correct to two decimal places. Hence find an approximate value for $k_{0}$.

Solution source

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){1/cos(#1*180/pi)};
    \def\xl{-0.5};
    \def\xu{pi/2+0.25};
    \def\yl{-0.5};
    \def\yu{5.2};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        
        % Draw a grid (optional)
        % \draw[grid] (-5,-3) grid (5,3);
        
        \draw[thick, blue, smooth, domain=\xl:{pi/2-0.1}, samples=100] 
            plot (\x, {\functionf(\x)});
        \draw[thick, red, smooth, domain=\xl:{pi/2-0.01}, samples=100] 
            plot (\x, {ln(2*\functionf(\x))});

        \draw[dotted] (0,0) -- ({pi/2},1);
        % \draw[dotted] (0,0) -- ({pi/2},2);
        \draw[dotted] (0,0) -- ({pi/2},{1.2981*pi/2});
        \draw[dotted] (0,0) -- ({pi/2},5);

        \draw[thick, dashed] ({pi/2},\yl) -- ({pi/2},\yu);
        \draw[thick, dashed] ({pi},\yl) -- ({pi},\yu);
        \node[below left] at ({pi/2},0) {$\pi/2$};
        \node[below left] at ({pi},0) {$\pi$};
    \end{scope}
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    \end{tikzpicture}
\end{center}

The red line is $y = \ln (2 \sec x)$, blue is $y = \sec x$.


We can see that if the gradient is too small it never touches the red line. If it is large it will cross the red line twice in that interval. For some value it will be perfectly tangent.

Since the line is tangent we must have 

\begin{align*}
&& y &= \ln (2 \sec x) \\
\Rightarrow && \frac{\d y}{ \d x} &= \frac{1}{2 \sec x} \cdot 2\sec x \tan x \\
&&&= \tan x \\
\Rightarrow && k_0 &=\tan x_0 \\
\Rightarrow && k_0 x_0 &= \ln(2 \sec x_0 ) \\
\Rightarrow && x_0 &= \cot x_0 \ln (2 \sec x_0)
\end{align*}

If $f(x) =x- \cot x \ln (2 \sec x)$, then $f'(x) =1 - 1+\ln(2\sec x) \cosec^2x  = \ln(2 \sec x)\cosec^2x $

so we should look at
\begin{align*}
x_{n+1} &= x_n - \frac{f(x_n)}{f'(x_n)} \\
&= x_n - \frac{x_n- \cot x_n \ln (2 \sec x_n)}{\ln(2 \sec x_n)\cosec^2x_n } \\
&= x_n \left (1 - \frac{\sin^2 x_n}{\ln (2 \sec x_n)}\right) +\sin x_n \cos x_n
\end{align*}

\begin{array}{c|c}
n & x_n \\ \hline 
1 & \frac{\pi}{4} \\
2 & 0.907701\ldots \\
3 & 0.91439340\ldots \\
4 & 0.914403867\ldots \\
5 & 0.91440386\ldots \\
6 & 0.91440386\ldots \\
\end{array}

The sign change test shows that $x_0 \approx 0.91$ and $k_0 = \tan(x_0) \approx 1.30$