1. Given equation

We need the number of distinct real roots of $x^7-7x-2=0.$

Let $f(x)=x^7-7x-2.$

We must determine how many distinct real solutions satisfy $f(x)=0$.

2. Use derivative to study the graph

Differentiate: $f'(x)=7x^6-7=7(x^6-1)=7(x^3-1)(x^3+1).$

So, $f'(x)=0 \iff x^6=1 \iff x=\pm 1$ for real $x$.

Thus the only critical points are at $x=-1$ and $x=1$.

Now check sign of $f'(x)$:

• If $|x|>1$, then $x^6>1$, so $f'(x)>0$.
• If $|x|<1$, then $x^6<1$, so $f'(x)<0$.

Hence:

• $f$ is increasing on $(-\infty,-1)$,
• decreasing on $(-1,1)$,
• increasing on $(1,\infty)$.

So the graph can cross the $x$-axis at most once in each of these three intervals.

3. Evaluate the function at critical points

At $x=-1$: $f(-1)=(-1)^7-7(-1)-2=-1+7-2=4.$

At $x=1$: $f(1)=1-7-2=-8.$

Also note end behavior: $\lim_{x\to -\infty} f(x)=-\infty, \qquad \lim_{x\to \infty} f(x)=\infty.$

4. Count roots interval-wise using IVT

(i) Interval $(-\infty,-1)$

• $f$ is increasing.
• $f(x)\to -\infty$ as $x\to -\infty$.
• $f(-1)=4>0$.

So by continuity, the function crosses zero exactly once in $(-\infty,-1)$.

(ii) Interval $(-1,1)$

• $f$ is decreasing.
• $f(-1)=4>0$ and $f(1)=-8<0$.

So by continuity, it crosses zero exactly once in $(-1,1)$.

(iii) Interval $(1,\infty)$

• $f$ is increasing.
• $f(1)=-8<0$ and $f(x)\to \infty$ as $x\to \infty$.

So it crosses zero exactly once in $(1,\infty)$.

5. Total number of distinct real roots

Therefore, the equation has $1+1+1=3$ distinct real roots.

So the correct option is $\boxed{\text{D: }3}.$

6. Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

They agree.