Wi: Working Rule: $$ x \sin x-1=0 $$

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1. We are given the equation  
   $$
   x \sin x - 1 = 0,
   $$  
   which can be rewritten as  
   $$
   x \sin x = 1.
   $$

2. Notice that the equation is nonlinear and cannot be solved exactly. We will therefore look for an approximate solution.

3. Define the function  
   $$
   f(x)= x \sin x - 1.
   $$
   We seek a value $$ x $$ such that $$ f(x)=0 $$.

4. First, we examine some values of $$ f(x) $$ to identify an interval where a sign change occurs (by the Intermediate Value Theorem):

- For $$ x=0 $$:  
     $$
     f(0)= 0 \cdot \sin 0 - 1 = -1.
     $$
     
   - For $$ x=\frac{\pi}{2}\approx 1.57 $$:  
     $$
     f\left(\frac{\pi}{2}\right)= \frac{\pi}{2} \sin\left(\frac{\pi}{2}\right) - 1 = \frac{\pi}{2} \cdot 1 - 1 \approx 1.57 - 1 = 0.57.
     $$
     
   Since $$ f(0) < 0 $$ and $$ f\left(\frac{\pi}{2}\right) > 0 $$, by the Intermediate Value Theorem there is at least one root in the interval $$ (0,1.57) $$.

5. To approximate the solution further, we evaluate $$ f(x) $$ at some intermediate values:

- For $$ x=1.1 $$:  
     We approximate $$\sin 1.1 \approx 0.891$$. Then  
     $$
     f(1.1)= 1.1 \times 0.891 - 1 \approx 0.98 - 1 = -0.02.
     $$
     
   - For $$ x=1.12 $$:  
     We approximate $$\sin 1.12 \approx 0.901$$. Then  
     $$
     f(1.12)= 1.12 \times 0.901 - 1 \approx 1.009 - 1 = 0.009.
     $$
     
   There is a sign change between $$ x=1.1 $$ (where $$ f(x) \approx -0.02 $$) and $$ x=1.12 $$ (where $$ f(x) \approx 0.009 $$). Hence, the solution lies between these values.

6. By further narrowing down within the interval and noting that the function is continuous, we can approximate the root to be around  
   $$
   x \approx 1.12.
   $$

Thus, the approximate solution to the equation  
$$
x \sin x - 1 = 0
$$
is  
$$
x \approx 1.12.
$$
The approximate solution to the equation $$ x \sin x - 1 = 0 $$ is $$ x \approx 1.12 $$.

Wi: Working Rule: \[ x \sin x-1=0 \]

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