Hence, x1 is a solution to the equation.

Using Calculus to Determine Uniqueness

While x1 is a solution, the question arises: can there be any other solutions? To address this, we can use calculus to analyze the function and prove that it can have only one solution.

Deriving and Analyzing the Function

Define the function:

fx 3 – 2√2x 32√2x - 6x

The derivative of fx is:

f'x ln(2√2)32√2x - ln(2√2)6x

simplifying further:

f'x ex [ ln(3) - 2ln(2√2) - ln(6) ]

Noting that ex is always positive, we focus on the term in the brackets:

ln (32√2/6) ln(9 - 8)/6 ln(1/6) -ln(6)

Since the term in the brackets is always negative, the derivative f'x is also negative for all x. Therefore, the function fx is always decreasing. A strictly decreasing function can have exactly one zero, and that zero must be at x1.

Alternative Methods for Solving the Equation

There are other methods to solve the equation, such as using inequalities and algebraic manipulation.

Using Inequalities

Consider the substitution:

a 3 - 2√2, b 32√2

Given that ab 6, the equation can be simplified to:

axbx abx, which is true only when x1.

Dividing by 3x

Another approach is to divide the entire equation by 3x:

1 - (2√2/3)x (32√2/3)x - 2x 0

Substitute cos(y) 2√2/3:

1 - cosx(y) (32/9x)cosx(y) - sin2x(y/2)cos2x(y/2) 1

This is true only when x1.

Computer Verification

The computer simulation verifies that x1 is the solution. Running the code or simulation will further confirm this observation.

Conclusion

Through various methods, we have determined that the solution to the equation 3 – 2√2x 32√2x - 6x 0 is x1. The unique solution can be proven using calculus and other algebraic techniques. The importance of understanding these methods extends to their applications in various scientific and engineering fields.