Two loudspeakers emit waves, find wavelength

[Moolisa]

I solved it, but then saw another solution online and am wondering if is is correct (since it is much faster than mine) and if my reasoning of it is correct

1. Homework Statement
Two loudspeakers emit sound waves along the x-axis. A listener in front of both speakers hears a maximum sound intensity when speaker 2 (s₂) at origin and speaker 1 (S₁) at x = 0.50 m. If s₁ slowly moved forward, the sound intensity decreases, then increases, reaching another maximum when s₁ is at x = 0.90 m.
a. What is wavelength

Homework Equations

Δ∅=2π (Δx/λ) +Δ∅₀ =(m) 2π

The Attempt at a Solution

My Solution[/B]
I used Δ∅=2π (Δx/λ) +Δ∅₀ =(m) 2π
I let m=1 when Δx=0.5m for equation 1 and m=0 when Δx=0.9m for equation 2. I manipulated they equations so each equaled Δ∅₀ and then set equations 1 and 2 equal to each other. Then I solved for λ=0.4m

Person Online
They just measured the two distances between maximum sound intensities in order to get wavelength. Now this makes sense to me (I think...) since it is essentially just measuring the distance between two crests. Is this reasoning wrong?More detailed version of my attempt
1st situation, m=1, Δx_a=0.5m
Δ∅=2π (Δx_a/λ) +Δ∅₀ =2π (1)
I manipulated the equation so
Δ∅₀=2π (1-(Δx_a/λ)) Equation 1

2nd Situation, m=0, Δx_b=0.9m
Δ∅=2π (Δx_b/λ) +Δ∅₀=2π (0)
Δ∅₀= -2π (Δx_b/λ) Equation 2

Then I set 1 and 2 equal to another and got wavelength

​

 

[RPinPA]

Your approach is correct. Look what happens when you manipulate those equations.
$$2\pi[1 - (\Delta x_a/\lambda) ]= -2\pi(\Delta x_b/\lambda)$$
$$1 - (\Delta x_a/\lambda) = -\Delta x_b/\lambda$$
$$1 = \frac {\Delta x_a - \Delta x_b} {\lambda}$$
$$\lambda = \Delta x_a - \Delta x_b$$

The difference between the two positions is one wavelength. Just as the other solution said.