Given, $\frac{d}{dx}\left(\frac{x·2^x-x}{1-\cos x}\right)=\left(\frac{x·2^x-x}{1-\cos x}\right)\left(f\left(x\right)+\log 2\right)$

$\Rightarrow \frac{\left(1-\cos x\right)\left[\frac{d}{dx}\left(x·2^x-x\right)\right]-\left(x·2^x-x\right)\frac{d}{dx}\left(1-\cos x\right)}{{\left(1-\cos x\right)}^2}=\left(\frac{x·2^x-x}{1-\cos x}\right)\left\{f\left(x\right)+\log 2\right\}$

$\Rightarrow \frac{\left(1-\cos x\right)\left(x·2^x\log 2+2^x-1\right)-\left(x·2^x-x\right)\left(\sin x\right)}{{\left(1-\cos x\right)}^2}=\left(\frac{x·2^x-x}{1-\cos x}\right)\left\{f\left(x\right)+\log 2\right\}$

$\Rightarrow \frac{\left(1-\cos x\right)\left(x·2^x\log 2+2^x-1\right)-\left(x·2^x-x\right)\left(\sin x\right)}{\left(1-\cos x\right)}=\left(x·2^x-x\right)\left\{f\left(x\right)+\log 2\right\}$

$\Rightarrow \frac{\left(1-\cos x\right)\left(x.2^x\log 2+2^x-1\right)-\left(x.2^x-x\right)\left(\sin x\right)}{\left(1-\cos x\right)\left(x.2^x-x\right)}-\log 2=f\left(x\right)$

$\Rightarrow \frac{\left(1-\cos x\right)\left(x·2^x\log 2+2^x-1\right)}{\left(1-\cos x\right)\left(x·2^x-x\right)}-\frac{\left(x·2^x-x\right)\left(\sin x\right)}{\left(1-\cos x\right)\left(x·2^x-x\right)}-\log 2=f\left(x\right)$

$\Rightarrow \frac{\left(x·2^x\log 2+2^x-1\right)}{\left(x·2^x-x\right)}-\frac{\left(x·2^x-x\right)\left(\sin x\right)}{\left(1-\cos x\right)\left(x·2^x-x\right)}-\log 2=f\left(x\right)$

$\Rightarrow \frac{\left(x·2^x\log 2+2^x-1\right)-\left(x·2^x-x\right)\log 2}{x\left(2^x-1\right)}-\frac{\left(\sin x\right)}{\left(1-\cos x\right)}=f\left(x\right)$

$\Rightarrow \frac{x·2^x\log 2-1+2^x-x·2^x\log 2+x\log 2}{x\left(2^x-1\right)}-\frac{\left(\sin x\right)}{\left(1-\cos x\right)}=f\left(x\right)$

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

$\therefore f\left(x\right)=\frac{1}{x}+\frac{\log 2}{2^x-1}-\frac{\sin x}{1-\cos x}$