Let $\sin(2x) = \frac{1}{7}$.  Find the numerical value of $\sin(x)\sin(x)\sin(x)\sin(x) + \cos(x)\cos(x)\cos(x)\cos(x)$.

$\text{(A) }\frac{2305}{2401}\qquad
\text{(B) }\frac{4610}{2401}\qquad
\text{(C) }\frac{2400}{2401}\qquad
\text{(D) }\frac{6915}{2401}\qquad
\text{(E) }\frac{1}{2401}\qquad
\text{(F) }0\qquad$

$\text{(G) }\frac{195}{196}\qquad
\text{(H) }\frac{195}{98}\qquad
\text{(I) }\frac{97}{98}\qquad
\text{(J) }\frac{1}{49}\qquad
\text{(K) }\frac{2}{49}\qquad
\text{(L) }\frac{48}{49}\qquad$

$\text{(M) }\frac{96}{49}\qquad
\text{(N) }\pi\qquad
\text{(O) }\text{none of the above}\qquad
\text{(P) }1\qquad 
\text{(Q) }2\qquad$
1. Given the identity for the double angle of sine:
   \[
   \sin(2x) = 2 \sin(x) \cos(x)
   \]
   We are given that:
   \[
   \sin(2x) = \frac{1}{7}
   \]
   Therefore:
   \[
   2 \sin(x) \cos(x) = \frac{1}{7}
   \]
   Solving for \(\sin(x) \cos(x)\):
   \[
   \sin(x) \cos(x) = \frac{1}{14}
   \]

2. We need to find the value of \(\sin^4(x) + \cos^4(x)\). We start by using the Pythagorean identity:
   \[
   \sin^2(x) + \cos^2(x) = 1
   \]
   Squaring both sides:
   \[
   (\sin^2(x) + \cos^2(x))^2 = 1^2
   \]
   Expanding the left-hand side:
   \[
   \sin^4(x) + 2 \sin^2(x) \cos^2(x) + \cos^4(x) = 1
   \]

3. We know from step 1 that:
   \[
   \sin(x) \cos(x) = \frac{1}{14}
   \]
   Therefore:
   \[
   \sin^2(x) \cos^2(x) = \left(\frac{1}{14}\right)^2 = \frac{1}{196}
   \]

4. Substitute \(\sin^2(x) \cos^2(x)\) into the expanded identity:
   \[
   \sin^4(x) + 2 \left(\frac{1}{196}\right) + \cos^4(x) = 1
   \]
   Simplify:
   \[
   \sin^4(x) + \cos^4(x) + \frac{2}{196} = 1
   \]
   \[
   \sin^4(x) + \cos^4(x) + \frac{1}{98} = 1
   \]

5. Solving for \(\sin^4(x) + \cos^4(x)\):
   \[
   \sin^4(x) + \cos^4(x) = 1 - \frac{1}{98}
   \]
   \[
   \sin^4(x) + \cos^4(x) = \frac{98}{98} - \frac{1}{98}
   \]
   \[
   \sin^4(x) + \cos^4(x) = \frac{97}{98}
   \]

The final answer is \(\boxed{\frac{97}{98}}\)