Find all solutions of 3cos(2x) = 2ln(x - 1).

Step-by-step solution

1. 1

   Fix the domain. The logarithm requires $x-1>0$, so every solution satisfies

   $x>1$

   As $x\to 1^{+}$ the right side tends to $-\infty$ while the left side stays in $[-3,3]$, so the difference $F(x)=3\cos 2x-2\ln(x-1)$ starts out large and positive.

2. 2

   Bound the search interval. The left side never exceeds $3$ in absolute value, so a solution needs $|2\ln(x-1)|\le 3$, that is

   $\ln(x-1)\le 1.5\;\Longrightarrow\;x\le 1+e^{1.5}\approx 5.4817$

   This converts an unbounded search into the finite interval $(1,\,5.4817]$ — the step that makes "find all solutions" answerable rather than open-ended.

3. 3

   Locate the sign changes of $F(x)=3\cos 2x-2\ln(x-1)$. Scanning the interval, $F$ changes sign exactly three times:

   $F(1.2)>0>F(1.35),\qquad F(2.4)<0<F(2.6),\qquad F(3.5)>0>F(3.7)$

   Between $3.7$ and $5.48$ the logarithm already dominates and $F$ stays negative, so there is no fourth crossing.

4. 4

   Refine each root by bisection. Halving each bracketing interval repeatedly gives

   $x_1\approx 1.283776,\qquad x_2\approx 2.490960,\qquad x_3\approx 3.584299$

5. 5

   Verify each root by substitution. Evaluating $F$ at the three values returns $-7.6\times 10^{-7}$, $-2.2\times 10^{-6}$ and $+1.6\times 10^{-6}$ — zero to within the rounding of six-decimal inputs $\checkmark$. For contrast, the nearby values $1.41538$, $2.40898$ and $3.55226$ give $F=-1.099$, $-0.370$ and $+0.170$, nowhere near zero, so they are not solutions.