Given,
x7−7x−2=0
x7−7x=2
Now taking f(x)=x7−7x (odd) & y=2
Now differentiating f(x) w.r.t x we get,
f′(x)=7(x6−1)=7(x2−1)(x4+x2+1)
f′(x)=0⇒x=±1
So points of extrema are -1&1, Now plotting the graph to check number of solution we get,

f(x)=2 has 3 real distinct solution.